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Elementary  Mechanics 

For  Engineering  Students 


By 
F.  M.  HARTMANN 

Professor  of  Physics,  Electrical  and  Mechanical  Engineering, 
Cooper   Union 


FIRST    EDITION 

FIRST   THOUSAND 


NEW  YORK: 
JOHN  WILEY  &  SONS. 

LONDON:     CHAPMAN  &  HALL,  LIMITED 
1910 


Copyright,  1910 
By  F.  M.  Hartmann 


Printed  by  Publishers  Printing  Co.,  New  York 


PREFACE 

THIS  little  volume  is  the  outcome  of  a  series  of 
lectures  on  elementary  mechanics,  delivered  to  the 
students  in  the  engineering  courses  of  the  Cooper 
Union  Day-  and  Night  Schools,  with  the  intention  of 
partially  clearing  the  ground  for  some  of  the  engineer- 
ing subjects  of  the  subsequent  years. 

In  all  cases,  some  knowledge  of  trigonometry  and 
elementary  physics  has  been  assumed;  and  it  is 
further  assumed  that  the  instructor  using  the  text  is 
thoroughly  familiar  with  the  students'  pretraining  in 
these  branches.  An  attempt  has  been  made  to  de- 
velop the  formulae  from  the  most  fundamental  prin- 
ciples; it  being  the  opinion  of  the  author  that  in  this 
way  the  student  derives  the  greatest  amount  of  good. 
No  excuse  is  made,  and  none  need  be  made,  for  em- 
ploying the  poundal  as  the  unit  force,  and  g  as  a 
proportionality  factor. 

A  short  chapter  on  the  determination  of  maxima 
and  minima  values,  by  algebraic  and  trigonometric 
methods,  is  given  with  the  hope  that  it  will  prove 
useful  in  solving  problems,  and  also  arouse  an  interest 
in  mathematical  analysis.  It  is,  of  course,  understood 
that  for  any  comprehensive  treatment  of  either  applied 
or  theoretical  mechanics,  the  application  of  the  differ- 


IV  PREFACE 

ential  and  integral  calculus  is  absolutely  necessary; 
but,  in  engineering  schools,  as  a  matter  of  economy 
in  time,  considerable  physics  must  be  taught  before 
the  student  is  familiar  with  the  methods  of  the  cal- 
culus. However,  were  it  not  a  matter  of  economy 
it  would  still  be  desirable;  for,  it  is  undoubtedly  true 
that  those  who  have  a  fair  knowledge  of  physical 
phenomena  acquire  the  calculus  more  readily.  The 
author  is  far  from  being  in  sympathy  with  those  en- 
gineers who  attempt  to  belittle  the  value  of  the  cal- 
culus. When  it  is  remembered  that  most  engineering 
problems  resolve  themselves  into  the  determination  of 
maxima  and  minima  values,  and  that  so  many  of  the 
factors  entering  into  a  problem  are  the  ratios  of 
variable  quantities,  it  follows  that  those,  other  things 
being  equal,  whose  minds  are  best  equipped  to  deal 
with  the  mathematical  relations  of  the  quantities  in- 
volved, will  do  the  most  efficient  work. 

Thanks   are   here   expressed   to   Messrs.    Riedel   and 
Bateman  for  supplying  the  problems. 

F.    M.    H. 

COOPER  UNION,  March,  1910. 


CONTENTS 

CHAPTER  I 

MOTION 

PAGE 

RECTILINEAR  MOTION  or  A  PARTICLE          .        .  2 

TRANSLATION  AND  ROTATION        ....  2 

UNIFORM  MOTION 2 

SPEED  AND  VELOCITY            3 

UNIT  VELOCITY              3 

VARIED  MOTION            4 

UNIFORMLY  VARIED  MOTION         ....  4 

ACCELERATION                4 

ANGULAR  MEASUREMENTS              ....  4 

RADIAN 4 

CURVATURE            4 

ANGULAR  VELOCITY 5 

PERIOD 5 

ANGULAR  ACCELERATION 7 

DEVELOPMENT    OF    FORMULAE    FOR    UNIFORMLY 

VARIED    MOTION  7 


VI  CONTENTS 

CHAPTER  II 

COMPOSITION   AND   RESOLUTION   OF 
MOTIONS 

PAGE 

COMPOSITION   OF   DISPLACEMENTS         .        .        .11 

COMPOSITION   OF   VELOCITIES        ....  13 

COMPOSITION   OF   ACCELERATIONS          ...  14 
RESOLUTION   OF    DISPLACEMENT,    VELOCITY   AND 

ACCELERATION 15 

TRIGONOMETRIC  ADDITION  OF  VECTORS       .        .  16 
CENTRIPETAL  ACCELERATION        .        .        .        -17 

SIMPLE    HARMONIC    MOTION         ....  20 

CHAPTER    III 

FORCE  AND  FRICTION. 
WORK  AND  ENERGY 

INERTIA 23 

FORCE            23 

MASS              24 

DENSITY 24 

COMPOSITION  AND  RESOLUTION  OF  FORCES        .  26 

FRICTION 27 

COEFFICIENT  OF  FRICTION 28 

ANGLE   OF   REPOSE 28 

CONE  OF  FRICTION 29 

WORK             30 

ENERGY — KINETIC  AND  POTENTIAL      .         .        •  '  32 

CONSERVATION    OF    ENERGY                  .        .  32 


CONTENTS 


Vil 


CHAPTER   IV 

MOMENTUM,      PRINCIPLE      OF      MOMENTS, 
AND   IMPACT 

PAGE 

MOMENTUM  

IMPULSE 

THREE    LAWS    OF    MOTION 

STRESS  

PRINCIPLE  OF  MOMENTS       .... 
MOMENT    OF    A    FORCE 

COUPLE,   MOMENT  OF 

CENTRE    OF    MASS 

CENTRE    OF    GRAVITY 41 

IMPACT  AND  MOMENTUM 41 

COEFFICIENT  OF  RESTITUTION  ....  43 
CONSERVATION  OF  MOMENTUM  ....  43 
Loss  OF  ENERGY  DURING  IMPACT  ...  46 


35 
35 
35 
36 
36 
37 
38 
39 


CHAPTER  V 

MOMENT    OF    INERTIA 

DEFINITION  OF 51 

RELATION  TO  TORQUE  AND  ANGULAR  ACCELERA- 
TION   .51 

RELATION  TO  KINETIC  ENERGY  AND  ANGULAR 

VELOCITY  .  .  .  .  .  .  .52 

CHANGE  OF   AXES 53 


RADIUS    OF    GYRATION 


59 


Vlli  CONTENTS 

CHAPTER   VI 

POWER   AND   ANGULAR   MOMENTUM 

PAGE 

POWER  EQUAL  TO  FORCE  TIMES  SPEED  .  .  60 
POWER  EQUAL  TO'  TORQUE  TIMES  ANGULAR 

VELOCITY 60 

ANGULAR  MOMENTUM 62 

POWER  EQUAL  TO  RATE  OF  CHANGE  OF  ANGULAR 

MOMENTUM  TIMES  ANGULAR  VELOCITY  62 


CHAPTER   VII 

TENSION    IN    CORDS 

TENSION  AND  ACCELERATION        .... 

ATWOOD'S  MACHINE  

INERTIA  OF  DRUM 

EQUIVALENT  MASS  OF  DRUM       .... 

CHAPTER   VIII 

MAXIMA   AND   MINIMA 

PRODUCT    OF    SINE   AND    COSINE         .        .        .71 
PRODUCT   OF   Two   VARIABLES,    WHOSE   SUM   is 

CONSTANT  72 


CONTENTS  IX 

PAGE 

SUM  OF  Two  VARIABLES,  WHOSE  PRODUCT  is 

CONSTANT 73 

MINIMUM  DISTANCE  BETWEEN  Two  MOVING 

PARTICLES,  WHOSE  PATHS  INTERSECT  .  74 

DIRECTION  OF  MOTION,  SUCH  THAT  A  GIVEN 
PARTICLE  MAY  MOVE  WITH  MINIMUM  SPEED 

AND  MEET  ANOTHER  PARTICLE  MOVING  IN 

• 

A  FIXED  PATH 75 

LEAST  FORCE  TO  MOVE  A  BODY  UP  AN  IN- 
CLINED PLANE 76 

PITCH  OF  THREAD  OF  JACK-SCREW        .        .        .78 
SET  OF  SAIL  WITH  RESPECT  TO  THE  DIRECTION 

OF  WIND  AND  MOTION  OF  VESSEL  80 


CHAPTER   IX 

PENDULAR    MOTION 

VIBRATION  AND  OSCILLATION       ....  83 

SIMPLE   PENDULUM 83 

PHYSICAL  PENDULUM  .        .  .        -85 

KATER'S  PENDULUM 87 

BALLISTIC  PENDULUM            .....  88 

GRAPHICAL  REPRESENTATION  OP  PENDULUM        .  90 

MINIMUM  LENGTH  OF   PENDULUM       ...  93 

EQUIVALENT  MASS  OF  PENDULUM       ...  93 

CONICAL  PENDULUM 95 


:  CONTENTS 

CHAPTER   X 

FALLING  BODIES  AND   PROJECTILES 


PAGE 


FORMULAE  FOR  HEIGHT,  TIME,  AND  VELOCITY     .  97 

RANGE  AND  TIME  OF  FLIGHT       ....  98 

INSTANTANEOUS    VALUE    OF    VELOCITY         .  99 

EQUATION  OF  PATH  OF  PROJECTILE        .        .  .99 


CHAPTER   XI 

ELASTICITY 

STRESS  AND  STRAIN 100 

HOOKE'S  LAW  AND  ELASTIC  LIMIT  .  .  .  100 

MODULUS  OF  TR ACTION AL  ELASTICITY  .  .  101 

ELASTICITY  OF  TORSION 102 

MODULUS  OF  RIGIDITY 102 

TORSION  AL  VIBRATION 103 

ELASTICITY  OF  FLEXURE 105 

NEUTRAL  PLANE  106 

NEUTRAL  Axis 107 

MOMENT  OF  AREA 109 

DEFLECTION  OF  RECTANGULAR  BAR  CLAMPED  AT 

ONE  END in 

BENDING  MOMENT 112 

DEFLECTION  OF  RECTANGULAR  BAR  SUPPORTED 

AT  ENDS,  AND  FORCE  APPLIED  AT  MIDDLE  114 


CONTENTS  XI 

PAGE 

ELASTICITY   OF   VOLUME 116 

LIQUIDS 116 

GASES  117 

CHAPTER   XII 

STATICS 

THREE  FORCES  MEETING  IN  A  POINT,  AND  PRIN- 
CIPLE OF  MOMENTS 120 

FORCE    POLYGON 121 

FUNICULAR    POLYGON 122 

RAY  POLYGON 124 

DETERMINING  POSITION  OF  EQUILIBRANT    .        .125 
PARALLEL  FORCES         .        .        .     *  .        .        .127 
UNIFORM    HORIZONTAL   LOADING          .        .        .128 
PARABOLA      ....        .        .        .        -131 

EXAMPLE    OF    THREE    FORCES    MEETING    IN    A 

POINT  .132 

PROBLEMS .137 


OF    THE 

UNIVERSITY 

OF 


ELEMENTARY  MECHANICS 

FOR 

ENGINEERING  STUDENTS 


CHAPTER    I 

MOTION 

OUR  concept  of  motion  is  the  relative  displacements 
that  occur  among  bodies;  but  to  describe  motion  it 
becomes  necessary  to  do  so  with  reference  to  some 
body  assumed  at  rest — it '  being,  of  course,  understood 
that,  so  far  as  we  know,  no  body  is  actually  at 
rest.  The  earth  together  with  the  other  planets  of  the 
solar  system,  and  their  attendant  satellites,  are  con- 
tinually changing  their  relative  positions;  and  further- 
more, the  sun  is  continually  changing  its  position  with 
respect  to  the  so-called  "fixed  stars,"  and  these  stars 
again  suffer  displacement  among  themselves. 

A  body  is  said  to  be  in  motion  when  it  occupies 
different  positions  at  different  intervals  of  time;  or, 
what  amounts  to  the  same  thing,  when  it  is  continu- 
ally changing  its  position.  In  terrestrial  mechanics 
the  surface  of  the  earth  is  assumed  at  rest,  and 


2  ELEMENTARY     MECHANICS 

bodies  having  a  fixed  position  on  the  earth's  surface 
are  said  to  be  at  rest.  Bodies  that  are  continually 
changing  their  positions,  relatively  to  the  earth's  sur- 
face, are  said  to  be  in  motion. 

Rectilinear  Motion.  In  our  preliminary  definitions, 
we  will  consider  the  motion  of  a  point  on  the  body 
rather  than  the  motion  of  the  body,  or  consider  the 
body  of  such  dimensions  that  all  parts  of  it  may  be 
considered  as  having  precisely  the  same  motion.  Such 
a  body  is  called  a  particle.  To  illustrate  this,  suppose 
some  body  to  move  so  that  a  point  in  the  body  is 
always  on  the  same  straight  line,  this  point  then  is 
said  to  have  rectilinear  motion.  The  body  as  a  whole 
may,  however,  not  have  a  rectilinear  motion;  for,  it 
may  at  the  same  time  be  spinning  about  a  line 
through  this  point  as  an  axis.  The  body  would  have 
then  both  motion  of  translation  and  rotation.  It  is 
obvious  that  we  are,  at  this  part  of  the  subject,  not 
prepared  to  deal  with  combinations  of  this  kind. 

Uniform  Motion.  A  body  passing  over  equal  dis- 
tances in  any  arbitrary  equal  intervals  of  time  is  said 
to  have  uniform  motion. 

It  must  be  remembered  that  the  statement,  "  uni- 
form motion  is  a  motion  such  that  equal  distances 
are  passed  over  in  equal  intervals  of  time,"  does  not 
define  uniform  motion.  For,  this  definition,  though  it 
includes  uniform  motion,  does  not  exclude  some  other 
types  of  motion  which  are  not  uniform.  This  may 


MOTION  3 

perhaps  be  best  illustrated  by  reference  to  a  seconds 
pendulum.  If  we  choose  for  our  points  of  reference 
the  extremities  of  its  path  and  for  the  interval  of 
time  the  second,  then  we  have  a  case  of  equal  dis- 
tances passed  over  in  equal  intervals  of  time.  This  is 
equally  true  if  one  half  second  be  taken  for  the  in- 
terval of  time.  .  But  the  pendulum  has  by  no  means 
a  uniform  motion;  for  its  motion  is  a  maximum  at 
the  middle  of  its  swing  and  at  either  extremity  of  its 
path  its  motion  is  momentarily  zero. 

Speed  and  Velocity.  By  speed  is  understood  the 
time  rate  of  motion.  There  is,  however,  another  ele- 
ment to  be  considered,  viz.:  direction.  When  both 
speed  and  direction  are  specified  it  is  called  Velocity. 
Velocity  is  defined  the  same  as  speed;  but  it  is  speci- 
fied with  respect  to  the  direction  of  the  motion,  i.e., 
it  is  a  directed  quantity.  Such  a  quantity  is  called  a 
Vector.  It  is  to  be  remembered  that  speed  is  purely 
a  ratio;  i.e.,  the  ratio  of  the  distance  passed  over  to 
the  time  consumed  in  passing  over  that  distance. 

Unit  Velocity.  Unit  velocity  is  a  velocity  such  that 
unit  distance  is  passed  over  in  unit  time.  In  the 
c.g.s.  (centimeter,  gram,  second)  system,  unit  velocity 
is  a  velocity  such  that  one  centimeter  is  passed  over 
per  second. 

In  the  F.P.S.  (foot,  pound,  second)  system,  unit 
velocity  is  a  velocity  such  that  one  foot  is  passed  over 
per  second. 


4  ELEMENTARY     MECHANICS 

Varied  Motion.  When  the  distances  passed  over  in 
equal  intervals  of  time  are  not  equal,  the  motion  is 
said  to  be  varied. 

Uniformly  Varied  Motion,  When  the  velocity 
changes  at  a  uniform  rate,  the  motion  is  said  to  be 
uniformly  varied;  and  the  particle  is  said  to  have  a 
constant  acceleration.  Acceleration,  then,  is  rate  of 
change  of  velocity.  When  the  velocity  of  a  particle 
is  increasing,  the  acceleration  is  positive,  when  de- 
creasing negative,  and  when  it  is  neither  increasing 
nor  decreasing,  i.e.,  when  the  motion  is  uniform,  the 
acceleration  is  zero. 

In  the  c.g.s.  system,  unit  acceleration  is  an  accelera- 
tion such  that  the  velocity  changes  at  the  rate  of 
one  cm.  per  sec.  per  sec.  In  the  F.P.S.  system,  unit 
acceleration  is  an  acceleration  such  that  the  velocity 
changes  at  the  rate  of  one  foot  per  sec.  per  sec. 

Angular    Measurements    and   Curvature   o]   a    Curve. 

Radian.  The  unit  of  angular  measure  employed  in 
mechanics  is  the  Radian,  and  is  the  angle  subtended 
by  an  arc  equal  to  the  radius.  In  any  case,  the 
measure  of  the  angle  is  the  ratio  of  the  length  of  the 
arc  to  the  radium  The  length  of  the  circumference 
of  a  circle  being  2  *r,  it  follows  from  the  definition 
that  the  sum  of  four  right  angles  is  equal  to  2  n 
radians. 

Curvature.     The   ratio    of   the    change    in    direction, 


MOTION  5 

measured  in  radians,  to  the  change  in  length  of  a 
curve  is  a  measure  of  the  curvature. 

To  illustrate  this:  The  direction  of  a  curve  at  any 
point  is  the  direction  of  the  tangent  to  the  curve  at 
that  point.  If  we  take  a  second  point  indefinitely 
close  to  the  first  point  and  draw  a  second  tangent, 
then  the  external  angle  between  these  two  tangents 
measures  the  change  in  direction;  this,  divided  by  the 
length  of  the  curve,  between  the  two  points,  is  a 
measure  of  the  curvature  of  the  curve. 

If  the  curve  is  the  circle,  then  this  ratio  is  a 
constant;  for  the  external  angle  between  the  tangents 
is  equal  to  the  angle  at  the  centre,  and  the  length 
of  arc  being  proportional  to  the  angle  at  the  centre, 
it  follows  that  the  ratio  of  change  in  direction  to 
change  in  length  is  a  constant. 

The  change  in  direction  in  going  once  round  a 
circle  is  four  right  angles  or  2  TT  radians.  The  length 

passed  over  is  2  nr.      Hence,  the  curvature  is  =  -. 

2  nr      r 

That  is,  the  curvature  of  a  curve  is  the  reciprocal 
of  the  radius.  The  curvature  of  a  straight  line  is 
zero,  and  the  radius  of  curvature  is  infinity. 

Angular  Velocity. — Consider  any  plane  figure  rotat- 
ing at  a  constant  rate  about  an  axis,  normal  to  the 
plane  of  the  figure,  with  a  period  T\  i.e.,  T  being 
the  time  of  one  rotation.  Any  point  on  the  plane 
will  then  be  moving  at  a  constant  speed;  but,  the 


6  ELEMENTARY     MECHANICS 

direction  of  motion  will  continually  change  at  a 
uniform  rate.  In  making  one  rotation,  or  what 
amounts  to  the  same  thing,  during  one  period,  the 
motion  changes  in  direction  by  an  amount  equal  to 
2  n  radians.  This  change  in  direction,  divided  by  the 
time  consumed  during  the  change,  is  a  measure  of 
the  angular  velocity  of  the  point.  That  is,  the  an- 
gular velocity  is  numerically  equal  to  -^,  and  is  rep- 
resented by  the  Greek  letter  w.  It  will  be  observed 
that  in  the  foregoing  discussion,  the  distance  of  the 
point  from  the  axis  was  not  considered,  and  that  the 
same  result  would  have  been  obtained  for  any  point. 
It  therefore  follows  that  all  points  on  the  body  have 
the  same  angular  velocity.  Since,  when  a  point  is 
moving  in  the  circumference  of  a  circle,  the  direction 
of  its  motion  at  any  point  is  in  the  direction  of 
the  tangent  to  the  circle  at  that  point,  and  further, 
since  the  radius  drawn  to  the  point  changes  in 
direction  at  the  same  rate  that  the  tangent  does,  it 
follows  that  the  angular  velocity  of  a  point  is  also 
measured  by  the  rate  of  change  of  direction  of  the 
radius.  We  may  then  define  angular  velocity  in 
general  as  the  ratio  of  the  angle  swept  out  by  a 
radius  to  the  time  consumed. 

Since  the  circumference  of  a  circle  is  equal  to 
2  Trr,  it  follows  that  if  a  point  is  moving  at  a  uniform 
rate  in  the  circumference  of  a  circle  with  a  period 


MOTION  7 

r,  its  linear  velocity  is  -=-.  That  is,  the  linear  veloc- 
ity is  equal  to  the  product  of  angular  velocity  and 
radius. 

Angular  velocity  may  be  uniform  or  variable.  When 
the  radius  sweeps  out  equal  angles  in  any  arbitrary 
equal  intervals  of  time,  the  rotating  body  has  a 
canstant  angular  velocity. 

When  the  angular  velocity  changes  at  a  uniform 
rate,  the  rotating  body  has  a  constant  angular  accel- 
eration. Angular  acceleration  has  precisely  the  same 
relation  to  angular  velocity  that  linear  acceleration 
has  to  linear  velocity.  That  is,  angular  acceleration 
is  the  rate  of  change  of  angular  velocity. 

We  are  now  prepared  to  state  some  of  these  re- 
lations symbolically. 

From  the  definition  of  uniform  motion,^ie  have 


distance  traveled 

velocity  =  -77—  -  y-  /  or  v 

time  consumed 


From  which 


s  =  vt.  |  V 

1! 


\ 

In   the    same    manner    for    constant   acceleration; 

change  in  velocity 
acceleration  =     ..         -  -    ,     ; 
time  consumed 


/ 


w 


8  ELEMENTARY     MECHANICS 

where  vt  is  the   initial  and  v2  the  final  velocity,  and 
a   the    acceleration. 

The  mean  velocity  of  a  body  having  a  uniformly 
accelerated  motion  is  obviously  the  half  sum  of  the 
initial  and  final  velocity;  i.  e., 

.V  =  —  ' 


2 

The    distance    passed    over    is    numerically    equal    to 
the  product  of  mean  velocity  and   time.     In  symbols 

•-W-^tSi.     .,.(.) 

Multiplying    equation    (i)    by    (2)    and    clearing    of 
fractions,  we  have 

2  a  s  =  v\  -  v\ (3) 

Again,    if   in   equation    (2)    we   substitute   for   v2   its 
value,  Vj_  +  at,  we  obtain 

2v,  +  at  at2 

s  =  -  -  /  =  v t  /  +    -.      .      .      (4) 

2  2 

If  the   initial  velocity   is  zero;    i.e.,   the   body   start 
from  rest,   we   have   the   following  equations: 

v  =  at, (5) 

vt 
s--,  .      (6) 

.-£ (7) 


-£ (8) 


MOTION 


The  equations  for  angular   motion  are   deduced   in 
a    precisely    similar    manner. 

By   definition   for   constant   angular   velocity; 


angle  swept  out 

angular  velocity  —  •—• 3, 

time   consumed 


or  in  symbols 


=  — ;   and  <p  =  at  t. 


In  the  same  manner,  for  constant  angular  acceleration; 

change  in  ang.   vel. 

angular  acceleration  =  --  ~~    -  =  —  -  —  , 

time  consumed 

or 

(U2   -    0)l 

a  =         t     *;      ....       (9) 

where  aj^   is  the  initial,  co2  the  final  angular  velocity, 
and   a   the  angular  acceleration. 

The  mean  angular  *  velocity  for  constant  angular 
acceleration  is  equal  to  the  half  sum  of  the  initial 
and  final  angular  velocities;  i.e., 


where   wa   is   the   mean   angular   velocity.     The   angle 
swept  out  is  the  mean  angular  velocity  multiplied  by 

the  time,  i.e., 

ait  +  to2 
<f>  =  wa  t  =  -  /.  (10) 

Multiplying   equation    (9)    by    (10)    and    clearing   of 
fractions,   we  obtain 


10  ELEMENTARY    MECHANICS 

Again,   if  in  equation   (10)    we    substitute  for  a)2  its 
value,   col  +  a  /,   we  obtain 

2  wl  +  a  t                       at2 
<p  =  ^  =  oi1M .      .      .      (12) 

2  2 

If,   as    before,   we   assume    the    body    starting   frorp 
rest,  we  obtain  the  following  relations: 

"  =  «/, (13)' 

cot 

<P  ~-    — , (14) 


(id) 


Such  are  the  algebraic  equations  that  must  neces- 
sarily follow  from  the  definitions. 

As  was  previously  stated,  linear  velocity  equals 
angular  velocity  multiplied  by  the  radius,  or  v  =  wr; 
dividing  by  t  we  have 

a  =  a  r ;  hence  v  :  a  : :  to  :  a. 


CHAPTER   II 

COMPOSITION    AND    RESOLUTION    OF 
MOTIONS 

THE  direction  in  which  a  body  is  moving  at  any 
point  of  its  path  may  always  be  represented  by  a 
straight  line;  and  likewise,  the  distance  over  which 
a  body  has  passed  from  a  given  fixed  point  may  be 
represented  to  a  given  scale  by  the  length  of  a  line. 
In  Fig.  i,  let  a  particle  move  from  A  along  the  line 
A  b  to  the  point  B\  then  from  B  along  the  line  BC 
to  the  point  C.  Now,  the  displacement  of  the  particle 
from  the  point  A  is  pre- 
cisely the  same  as  though 
the  motion  had  taken 
place  along  the  line  A  C 
by  an  amount  equal  to 
the  length  of  the  line 
AC]  or  along  the  path  ADC.  That  is,  it  is  im- 
material, so  far  as  final  displacement  is  concerned, 
whether  the  particle  first  move  in  the  direction  A  B 
by  an  amount  A  B  and  then  in  the  direction  B  C 
by  an  amount  BC',  or,  first  move  in  the  direction 
A  D,  parallel  to  BC,  by  an  amount  BC  and  then  in 
the  direction  DC,  parallel  to  A  B,  by  an  amount 


12  ELEMENTARY    MECHANICS 

AB-,  or  whether  it  move  along  the  line  A  C  by  an 
amount  A  C.  Suppose  now,  the  particle  to  be  at 
the  point  A,  and  moving  at  a  uniform  rate,  rela- 
tively to  the  earth's  surface,  along  the  line  A  B, 
and  simultaneously  with  this  the  plane  surface  upon 
which  the  particle  is  moving,  to  move  at  a  uniform 
rate,  relatively  to  the  earth's  surface,  carrying  the 
particle  with  it,  in  the  direction  A  D.  Assume  that 
the  relative  rates  are  such  that  the  particle  moves  on 
the  surface  from  A  to  B,  while  the  surface  has 
moved  by  an  amount  equal  to  A  D.  The  particle 
will  then  be  situated  at  C,  and  the  displacement  will 
again  be  the  same  as  though  the  particle  had  moved 
along  the  line  A  C  from  A  to  C.  But  in  this  case 
the  particle  has  actually  moved,  relative  to  the  earth's 
surface,  along  the  line  A  C.  For,  take  some  fractional 
part  of  the  line  A  B,  such  as  A  b,  and  a  like  fractional 
part  of  the  line  A  D,  such  as  A  d.  Draw  the  lines 
d  c  and  b  c  respectively  parallel  to  A  B  and  A  D.  Then 
from  the  conditions,  while  the  particle  has  moved 
from  A  to  6,  the  surface  has  moved  the  amount  A  d 
parallel  to  A  D.  Hence,  the  point  c  gives  the  position 
of  the  particle  at  that  instant;  but  by  construction, 
the  triangles  A  d  c  and  ADC  are  similar,  hence  the 
point  c  falls  on  the  line  A  C\  and  this  is  true  no 
matter  what  fractional  part  of  A  B  is  taken;  hence 
the  path  of  the  particle  is  the  line  A  C.  It  must  be 
remembered  that  the  line  A  C  is  fixed  with  respect  to 


COMPOSITION   AND   RESOLUTION    OF   MOTIONS      13 

some  surface  assumed  at  rest,  such  as  the  earth's 
surface. 

If  now,  we  assume  the  diagram  drawn  to  such  a 
scale  that  the  lengths  A  B  and  A  D  represent  the  dis- 
tances passed  over  in  a  unit  of  time,  then  these  lengths 
also  represent  the  velocities  in  magnitude  and  direc- 
tion. And,  since  A  C  represents  both  in  magnitude 
and  direction  the  displacement  of  the  particle  in  a 
unit  of  time,  it  follows  that  it  also  represents  the 
resultant  of  the  velocities  A  B  and  A  D. 

If  the  velocities  be  variable,  and  at  any  instant  the 
velocity  in  the  direction  A  B  is  equal  to  A  B  and  at 
the  same  instant  the  velocity  in  the  direction  A  D  is 
equal  to  A  D,  then  A  C  represents  in  magnitude  and 
direction  the  instantaneous  resultant  velocity.  Since 
A  C  represents  the  resultant  velocity  when  the  two 
component  velocities  are  constant,  it  must  also  repre- 
sent the  resultant  velocity  at  any  instant;  for,  the 
velocities  at  the  instant  being  such  that  the  two  dis- 
placements A  B  and  A  D  would  be  produced  in  a  unit 
of  time,  and  the  resultant  of  these  two  displacements 
being  A  C,  which  would  also  have  been  produced  in 
unit  time,  it  is  obvious  that  A  C  represents  the  in- 
stantaneous resultant  velocity.  We  therefore  have  the 
following  theorem: 

The  resultant  of  two  concurrent  coplanar  velocities 
is  represented  in  magnitude  and  direction  by  the 
diagonal  of  the  parallelogram  constructed  upon  the 


14  ELEMENTARY    MECHANICS 

component  velocities  as  sides;  the  component  veloc- 
ities being  drawn  from  a  common  point  to  the  same 
scale  and  in  the  proper  directions. 

If  we  assume  constant  accelerations  in  the  two 
directions,  and  draw  our  diagram  to  such  a  scale  that 
A  B  and  A  D  represent  the  velocities  acquired  in  the 
two  directions  in  a  unit  of  time,  then  the  instanta- 
neous resultant  velocity  at  the  end  of  a  unit  of  time 
will  be  represented  by  A  C.  But,  since  this  velocity 
has  been  acquired  in  a  unit  of  time,  the  line  A  C  also 
represents  the  acceleration.  Accelerations  may  there- 
fore be  combined  by  the  same  methods  as  velocities. 

It  can  readily  be  seen  that  the  foregoing  demonstra- 
tions can  be  extended  to  any  number  of  coplanar 
components;  for,  it  is  only  necessary  to  find  the  re- 
sultant for  any  two  of  them,  combine  this  resultant 
with  a  third  component,  and  so  on,  until  the  final 
resultant  is  obtained;  the  final  resultant  being  in- 
dependent of  the  order  in  which  the  components  are 
taken. 

As  has  been  previously  stated,  quantities  in  which 
direction  is  specified,  as  well  as  magnitude,  are  vector 
quantities.  Displacements,  velocities,  and  accelera- 
tions are  vector  quantities.  In  Fig.  i,  the  lines  A  B, 
B  C,  and  A  C  represent  vectors.  By  referring  to  Fig. 
i,  it  will  be  seen  that  the  resultant  of  the  two  com- 
ponent vectors  may  also  be  obtained  by  laying  off 
the  vector  A  B,  then  from  the  terminal  B  laying  off 


COMPOSITION   AND   RESOLUTION    OF   MOTIONS      1$ 

the  vector  BC,  and  by  joining  A  and  C  we  obtain 
the  resultant.  Or  expressed  vectorially 

AB  +  BC  =  AC. 

The  student  must  consider  carefully  what  this  means. 
It,  of  course,  does  not  mean  that  the  length  of  the 
line  A  B  plus  the  length  of  the  line  B  C  equals  the 
length  of  the  line  AC.  But  it  does  mean  that  in 
effect,  as  regards  displacements,  velocities,  or  accelera- 
tions, the  sum  of  A  B  and  B  C  equals  A  C.  This 
addition  of  vectors  applies  to  any  number  of  coplanar 
vectors;  to  find  the  resultant  or  vector  sum,  it  is  only 
necessary  to  lay  off  any  vector  in  the  proper  direction 
and  to  proper  scale,  and  from  its  terminal  lay  off 
a  second  vector,  etc.  The  line  then  joining  the  be- 
ginning of  the  first  vector  and  the  terminal  of  the 
last  vector,  is  the  resultant,  or  vector  sum.  The 
vector  sum  of  a  number  of  vectors  is  independent  of 
the  order  of  addition. 

Resolution.  A  displacement,  a  velocity,  or  an  ac- 
celeration may  be  resolved  into  components  just  as 
well  as  components  may  be  combined.  If  in  Fig.  i, 
a  particle  start  from  A  and  move  to  C,  its  final  dis- 
placement is  independent  of  the  path  pursued  in  going 
from  A  to  C.  In  the  same  manner,  if  A  C  represents 
the  velocity  of  a  particle  at  any  instant,  then  since 
A  C  equals  the  components  A  B  plus  B  C  it  can  be 
resolved  into  these  two  components,  or  any  other  two 
components  whose  sum  equals  A  C.  But,  since  each 


i6 


ELEMENTARY   MECHANICS 


of  these  components  can  again  be  resolved  into  two 
components,  and  these  individual  components  can 
again  be  resolved,  it  follows  that  any  vector  can  be 
resolved  into  any  number  of  coplanar  vectors  whose 
sum  is  equal  to  the  given  vector. 

Trigonometric  Addition  of  Vectors.     The  resultant  of 
two  coplanar  vectors  is  readily  found  by  the  formula: 


FIG.  2. 

R  =  (A2  -I-  B*  +  2ABcos6)*-,  where  R  is  the  result- 
ant, A  and  B  the  two  component  vectors,  inclined  to 
each  other  by  the  angle  6.  The  formula  gives,  how- 
ever, the  magnitude  only,  and  not  the  direction.  To 
find  the  magnitude  and  direction  of  the  resultant  of 
two  or  any  number  of  coplanar  vectors,  proceed  as 
follows:  Let  in  Fig.  2,  A,  B,  C,  and  D  be  the  given 
vectors;  inclined  respectively  to  the  X  axis  by  the 


COMPOSITION   AND   RESOLUTION    OF    MOTIONS      I/ 

angles  <ply  <f>2,  <p3,  and  ^4.  Multiplying  each  vector 
by  the  sine  and  cosine  of  its  angle  of  inclination,  the 
vertical  components,  Ay,  By,  Cy,  and  Dy  are  found; 
and  likewise,  the  horizontal  components  Ax,  Bx,  Cx, 
and  Dx.  Taking  now,  the  algebraic  sum  of  the 
vertical  components,  Ey  is  found,  and  Ex  for  the 
algebraic  sum  of  the  horizontal  components.  The 
resultant,  E,  is  now  completely  specified,  it  being  the 
hypothenuse  of  the  right  triangle  whose  legs  are 
Ey  and  Er,  and  its  inclination  to  the  X  axis  is  given  by 


It  is  not  necessary  to  draw  the  diagram;  it  is  only 
necessary  to  multiply  each  vector  by  the  sine  and 
cosine  of  its  angle  of  inclination,  proper  attention 
being  given  to  the  signs  of  the  trigonometric  functions, 
then  taking  the  algebraic  sum  of  the  vertical  com- 
ponents and  of  the  horizontal  components,  the  two 
quantities  obtained  being  respectively  the  vertical  and 
horizontal  component  of  the  resultant. 

Centripetal  Acceleration.  Velocity  has  been  defined 
as  being  constant  when  the  rate  of  motion  —  speed- 
is  constant,  and  further,  when  there  is  no  change  in 
direction.  When  there  is  a  change  in  direction,  even 
though  the  speed  be  constant,  the  velocity  varies,  and 
hence  there  is  an  acceleration. 

A  particle  moving  with  a  uniform  speed  in  the  cir- 
cumference of  a  circle  is  such  a  case, 


i8 


ELEMENTARY   MECHANICS 


Let  a  particle  be  moving  in  the  circumference  of  a 
circle  whose  centre  is  O,  with  a  uniform  speed.     The 
speed     being     constant,     the     distance      travelled      is 
proportional  to  the  time.     Let  us  consider  the  change 
from    the    instant   the   body    is 
passing    through    the    point  a, 
Fig.    3,    with    a  velocity   v    in 
the  direction  af  tangent  to  the 
curve    at    the    point    a.      Had 
there  been  no  acceleration,  the 
body      would     have    continued 
to    travel    in  the    direction    af 
with  constant   speed;     but  due 


FIG.  3. 


to  an  acceleration,  it  is  continually  deviated  from 
the  straight  line  and  travels  along  the  curve.  By  the 
time  the  particle  has  reached  the  point  d,  it  has 
been  displaced  at  right  angles  from  its  path  at  a, 
a  distance  a  e.  Assume  the  distance  a  d,  along  the 
arc,  an  immeasurably  small  fractional  part  of  the 
circumference  of  the  circle,  then  ae  is  very  small 
compared  with  the  radius.  Under  the  assumed  con- 
ditions, the  distances  along  the  arc  and  chord  are 
sensibly  equal.  Also,  the  distance  a  e  along  the 
radius  will  be  so  small  that  the  acceleration,  dur- 
ing the  time  required  to  bring  about  this  displace- 
ment, is  sensibly  constant;  hence,  we  may  apply  the 
formula  for  uniformly  accelerated  motion.  By  equa- 
tion (8),  Chapter  I,  we  have  a  e  =  -— ;  where  t  is 


COMPOSITION    AND    RESOLUTION    OF    MOTIONS      IQ 


the  time  required  to  bring  about  the  change.  Further, 
we  have  a  d  =  v  t,  and  by  the  geometry  of  the  figure, 
we  have  the  following  proportion: 

ba:ad::ad\ae\  from  which,  by  substitution, 

at2 
2r:vt::vt:  — . 

2 

I/2 

Solving,   we   have  a  =  — ;    i.e.,  the  acceleration  along 

the  radius  is  equal  to  the  square  of  the  speed  multi- 
plied by  the  curvature  of  the  curve. 

It  is  to  be  remembered  that  distances  in  the  figure 
are  exaggerated,  and  that  the  assumptions  made  in 
the  demonstration  are  true  only  when  the  distance 
ad  becomes  indefinitely  small. 

This  formula  is  so  important  that  it  may  be 
of  value  to  deduce  it  by  an  entirely  independent 
method. 

In  Fig.  4,  let  be  represent  the  direction  and  mag- 
nitude of  the  velocity  of  the  par- 
ticle at  the  instant  it  is  passing 
through  the  point  b.  It  is  ob- 
vious that  as  the  radius  r  ro- 
tates about  the  point  O,  be  is 
always  at  right  angles  to  it;  hence 
it  may  be  represented  by  a  second 
line  o  d,  passing  through  the  center 
O,  parallel  to  be  and  equal  to  it  in  length.  As 
v  always  represents  in  direction  and  magnitude  the 


FIG.  4. 


20  ELEMENTARY   MECHANICS 

motion  of  the  terminal  of  the  radius  r,  so  does  a 
represent  the  motion  of  the  extremity  of  v.  Hence  we 
may  write,  since  the  angular  velocity  of  r  and  v  is 
the  same, 

r  :  v  :  :  v  :  a, 
or 

*-V-r (i) 

But  since  a  is  the  rate  of  motion  of  the  extremity 
of  v,  and  v  is  the  rate  of  motion  of  the  ex- 
tremity of  r  —  or  the  rate  of  motion  of  the  particle 
—it  follows  that  a  is  the  rate  of  a  rate  of  motion; 
i.e.,  an  acceleration,  and  the  theorem  is  proved. 

Simple  Harmonic  Motion.  S.  H.  M.  is  a  mo- 
tion along  a  fixed  path,  such  that  the  acceleration 
towards  a  fixed  point  in  the  path  is  always  propor- 
tional to  the  displacement,  measured  along  the  path, 
from  the  fixed  point. 

From  this  definition,  it  at  once  follows  that  no 
matter  what  the  velocity  of  the  particle,  when  passing 
through  the  fixed  point,  it  must  finally  come  to 
rest;  since  the  acceleration  is  towards  the  fixed  point. 
Furthermore,  after  coming  to  rest  the  particle  will  be- 
gin to  move  towards  the  fixed  point  with  increasing 
velocity,  and  by  the  time  it  again  reaches  this  point, 
will  have  a  velocity  precisely  equal  in  magnitude, 
but  opposite  in  direction  to  that  when  previously 
passing  through  the  point.  It  will  then  move  to 


COMPOSITION   AND   RESOLUTION   OF   MOTIONS       21 

an  equal  displacement  on  the  opposite  side  of  the 
fixed  point,  and  so  on.  In  other  words,  the  mo- 
tion is  periodic;  i.e.,  repeats  at  regular  intervals. 
The  fixed  point  is  called  the  position  of  equilib- 
rium', there  being  no  acceleration.  The  greatest 
distance  from  the  fixed  point,  measured  along  the 
path,  is  the  amplitude.  The  distance  at  any  in- 
stant from  the  fixed  point  is  the  displacement. 
The  time  required  to  pass  through  a  cycle;  ie., 
the  time  required  to  come  to  the  original  condi- 
tion, which  means  being  at  the  same  point  and 
going  in  the  same  direction,  is  a  period  and  repre- 
sented by  T. 

To  make  a  complete  general  solution  is  far  too 
complex  mathematically  for  elementary  mechanics; 
but  a  good  idea  may  be  obtained  by  studying  some 
particular  case  in  detail. 

Consider  a  particle,  Fig.   5,  moving  with  a  uniform 
speed   in   the   circumference   of  a 
circle   whose  centre  is  C   and  ra- 
dius r.    Let  the  period  be  T,  then 

2  71 

the   angular   velocity   is    —  =  o>; 

and    the  angle  at   any    instant    is 

<p  =  cot',  the  angle  and  time  being 

FIG.   5. 
both   measured    from    the    origin, 

O.  If  the  velocity,  V,  be  resolved  into  two  compo- 
nents, one  at  right  angles  and  the  other  parallel  to 


22  ELEMENTARY  MECHANICS 

the   diameter   OB,   the    parallel    component    will    be 

v  =  V  sin  <p  =  w  r  sin  cot;  .  .  -  .  (2) 
a>  and  r  being  constant,  it  follows  that  the  velocity 
parallel  to  the  diameter  is  proportional  to  the  sine 
of  the  angle.  From  equation  (i)  we  have  the  radial 

F2 
acceleration    equal    to    • — .     The    component    of    this 

acceleration,    parallel    to    the    diameter,    is 

F2 
/  =  -        cos   <p  =  aj2  r  cos  wt.      .       .      (3) 

But,    r    cos    cot    is    equal    to    the    displacement    s; 
hence, 

/="'*  =  ^ (4) 

That  is,  since  a>2  is  a  constant,  the  acceleration 
along  the  diameter  varies  directly  as  the  displacement 
from  the  center  of  the  circle.  Therefore,  when  a 
point  moves  with  a  constant  speed  in  the  circum- 
ference of  a  circle,  the  projection  of  this  motion  onto 
a  diameter  of  the  circle  is  a  simple  harmonic  motion. 
Students  cannot  become  too  familiar  with  these 
formulae.  There  are  no  equations  of  more  universal 
application  in  mechanics  and  physics  than  those  of 
simple  harmonic  motion. 


CHAPTER    III 

FORCE    AND   FRICTION 
WORK    AND    ENERGY 

Inertia  and  Force.  So  far,  motion  has  been  con- 
sidered in  the  abstract,  without  considering  the 
nature  of  the  body  moved. 

As  a  result  of  experience  we  know  that  it  requires 
an  effort  to  change  the  rate  of  motion  of  a  body; 
and  further,  that  the  intensity  of  the  effort  depends 
upon  the  body  moved  and  the  rate  at  which  the 
change  is  brought  about;  i.e.,  the  acceleration. 

Broadly  stated,  every  body  persists  in  maintaining 
whatever  rate  of  motion  it  may  happen  to  have; 
this  is  the  chief  characteristic  of  matter  and  is 
termed  inertia.  This  idea  must  not  be  confused 
with  inactivity. 

As  previously  stated,  an  effort  is  required  to  ac- 
celerate matter,  and  this  effort  is  termed  Force. 

We  may  then  define  force  as  that  which  changes 
or  tends  to  change  the  rate  of  motion  of  a  body. 

Suppose  we  have  two  spheres  of  equal  volume, 
but  of  different  materials;  say  one  of  wood  and 
the  other  of  lead,  lying  on  a  perfectly  smooth  hor- 

[23] 


24  ELEMENTARY   MECHANICS 

izontal  surface,  and  accelerate  them  equally,  it  will 
be  found  that  a  greater  force  is  required  for  the 
leaden  sphere  than  is  required  for  the  wooden  one. 
Were  they  of  the  same  material,  we  should  find 
the  forces  equal.  Were  they  of  the  same  material 
but  of  different  dimensions,  it  would  be  found  that 
the  greater  force  is  required  for  the  larger  sphere. 
It  is  evident  that  the  substances  being  the  same, 
the  larger  sphere  contains  the  greater  quantity  of 
matter.  That  is,  by  operating  on  bodies  composed 
of  the  same  material,  we  find  that  the  greater  the 
quantity  of  matter  the  greater  the  force.  In  the 
case  of  the  two  spheres  of  equal  volume,  one  of 
wood  and  the  other  of  lead,  it  is  found  that  the 
leaden  sphere  requires  a  greater  force  than  the 
wooden  sphere  when  they  are  given  equal  accelera- 
tions; hence  it  is  assumed  that  the  leaden  body  con- 
tains a  greater  quantity  of  matter.  The  quantity  of 
matter  a  body  contains  is  called  its  Mass. 
The  mass  per  unit  volume  is  the  Density;  i.e., 

Mass  r.      M 

Density  =  -=—. ,  or  D  =  — . 

Volume  V 

We  may  now  embody  some  of  the  foregoing  state- 
ments regarding  mass,  force,  and  acceleration  in 
the  form  of  equations.  That  is,  that  the  force  is 
proportional  to  the  mass  and  to  the  acceleration  it 
produces.  In  symbols, 

F  ^  ma. 


UNIVERSITY 

OF 


AND    FRICTION  —  WORK   AND   ENERGY      25 

By  introducing  a  constant  we  may  use  the  equality 
sign,  i.e., 

F  =  kma;     .....      (i) 

k  depending  upon  the  units  chosen,  and  may  be 
made  unity  by  choosing  the  proper  units.  Equation 
(i)  shows  that  the  mass  of  a  body  may  be  measured 
by  the  force  required  to  produce  a  given  accelera- 
tion; i.e.,  by  its  inertia. 

In  the  c.g.s.  system,  the  unit  mass  is  the  gram, 
the  unit  acceleration  is  that  when  the  velocity  changes 
at  the  rate  of  i  cm.  per  sec.  per  sec.  Hence  if 
we  choose  as  our  unit  force  that  force  which  will 
accelerate  the  mass  of  i  gram  i  cm.  per  sec.  per  sec., 
k  in  equation  (i)  becomes  unity.  This  unit  force 
is  called  the  dyne. 

In  the  F.P.S.  system,  the  unit  force  is  the 
poundal,  and  is  that  force  which  will  accelerate  the 
mass  of  i  pound  i  foot  per  sec.  per  sec.  There  is 
another  unit  in  common  use;  namely,  the  engineers' 
unit,  called  pound.  This  force  is  equal  to  the 
weight  of  a  pound,  and  hence  is  a  variable,  depend- 
ing upon  latitude  and  elevation.  In  any  case  it  is 
the  force  that  will  accelerate  the  mass  of  one  pound 
g  feet  per  sec.  per  sec.;  where  g  is  the  acceleration 
due  to  the  earth's  gravitational  field,  for  the  par- 
ticular locality.  In  this  case,  k  in  equation  (i) 

becomes  —  ;  i.e., 

o 


26  ELEMENTARY   MECHANICS 


In  most  cases  32.2  is  sufficiently  accurate  for  the 
value  of  g,  and  in  most  engineering  computations 
32.  may  be  used. 

Composition  and  Resolution  of  Forces.  Assume 
a  body,  whose  mass  is  m,  acted  upon  by  two  forces 
in  such  a  manner  that  the  two  accelerations  A  b 
and  A  d  (Fig.  6),  are  imparted.  The  resultant 
acceleration  will  be  represented  by  A  c.  The  force 


FIG.  6. 

required  to  produce  the  acceleration  A  b  is  equal  to 
m  X  A  b.  Let  this  be  represented  by  A  B.  Likewise, 
let  A  D  represent  to  the  same  scale  the  product  of 
m  and  A  d;  i.e.,  the  force  required  to  produce  the 
acceleration  A  d.  The  force  required  to  produce  the 
acceleration  A  c  is  equal  to  the  product  of  m  and 
A  c.  Draw  B  C  and  D  C  respectively  parallel  to  A  D 
and  A  B.  Since  A  b  and  A  d  are,  by  construction,  like 
fractional  parts  of  A  B  and  A  D,  the  point  c  must 


FORCE  AND   FRICTION — WORK  AND   ENERGY      27 

fall    on    the    line    A  C.     From    the    similarity    of    the 

figure  we  have 

AC       ABf 
Ac     =  Ab' 

but  — r  =  m    by    construction.      Therefore,    m  X  A  c 
A  b 

=  AC]  i.e.,  AC  represents  the  force  required  to  give 
the  acceleration  Ac  to  the  mass  m,  and  hence  is  the 
resultant  of  the  forces  A  B  and  A  D.  The  resultant 
of  two  or  more  concurrent  coplanar  forces  is  there- 
fore found  in  the  same  manner  that  the  resultant 
of  displacements,  velocities,  or  accelerations  is  found. 

From  what  has  been  shown  in  the  composition  of 
forces,  it  is  readily  seen,  that  a  force  may  be  resolved 
into  components  the  same  as  any  other  vector  quantity. 

Friction.  When  two  surfaces  are  in  contact  and 
are  caused  to  move  relatively  to  each  other,  a  re- 
sistance is  experienced,  which  is  a  function  of  the 
normal  pressure  between  the  surfaces  and  of  the 
rate  of  relative  motion.  This  resistance  is  called 
the  force  of  friction.  The  following  statements  are 
usually  given  as  being  approximately  true. 

(1)  The    force    of    friction    is    directly    proportional 
to   the  normal   pressure  between  the   surfaces. 

(2)  The    force    of    friction    for    any    given    normal 
pressure    is    independent    of   the   area   of   the   surfaces 
in  contact. 

(3)  The  force  of  friction  is  independent  of  the  rate 
of  motion. 


28  ELEMENTARY   MECHANICS 

The  first  and  second  statements  are  practically  true 
for  ordinary  pressures.  When  the  pressures  become 
high  there  is  a  very  wide  departure;  and  at  what 
pressure  this  takes  place  depends,  of  course,  on  the 
nature  of  the  substances  in  contact.  The  third  state- 
ment is  not  even  practically  true  for  most  cases.  In 
a  great  many  cases  when  the  rate  of  motion  varies 
only  slightly,  then  other  things  being  equal,  the  force 
of  friction  is  for  practical  purposes  constant.  But 
whenever  there  is  considerable  variation  in  the  rate 
of  motion  there  is  also  a  measurable  change  in  the 
force  of  friction. 

Coefficient  of  Friction.  The  ratio  of  the  force  of 
friction  to  the  normal  pressure  is  called  the  coefficient 
of  friction.  In  symbols 

F 

H  =  — ,  or  F  =  //  AT;        .      .      .      (3) 

where  ft  is  the  coefficient,  F  the  force,  and  N  the 
normal  pressure.  From  what  has  been  stated,  it  is 
clear  that  this  formula  can  be  used  only  when  the 
conditions  under  consideration  approximate  very  close- 
ly, as  regards  intensity  of  pressure  and  rate  of  motion, 
to  those  conditions  under  which  the  coefficient  of 
friction  was  determined. 

Angle  of  Repose.  If  a  plane,  having  a  body  resting 
upon  it,  be  inclined  so  that  the  body  is  just  on  the 
point  of  sliding,  the  angle  which  the  plane  makes 
with  the  horizontal  is  called  the  angle  of  repose.  It 


FORCE  AND   FRICTION — WORK  AND   ENERGY     2Q 


will  now   be   shown  that   the   coefficient   of   friction   is 
equal  to   the  tangent   of  the   angle  of  repose. 

Let,  in  Fig.  7,  the  plane  be 
inclined  at  an  angle  p  such 
that  the  body,  whose  weight  is 
W,  is  just  on  the  point  of  slid- 
ing. Resolving  the  weight  into 


FIG.   7. 


two  components,  we    obtain   for   the   force   parallel  to 
the  plane,   which   is   equal   to   the   force   of  friction, 

p  =  W  sin  p, 
and    for    the    normal    pressure 

R  =  W  cos  p. 

By    definition,    the    coefficient    of    friction    is   the    ratio 
of  the  force  of  friction  to  the  normal  pressure;    hence 

W  sinp 


=  tan  p. 


(4) 


W  cos  p 

Assume,  as  in  Fig.  8,  a  body  of  weight  W  lying 
upon  a  horizontal  surface  and  having  applied  a  force 
F  making  an  angle  (f>  with  a  normal 
to  the  plane.  The  normal  pressure 
between  the  two  surfaces  then  is 

R  =  W  +  F  cos  y, 

and  the  force  tending  to  move  the 
body  is 

/  =  F  sin  <p. 

Now,  for  sliding  to  be  impending,  the  force  parallel 
to  the  surface  must  be  equal  to  the  product  of  the 
coefficient  of  friction  and  the  normal  pressure;  hence, 


FIG.  8. 


30  ELEMENTARY   MECHANICS 

for  the  body  to  be  on  the  point  of  sliding,  we  must 
have 

F  sin  <p  =  fi  (W  +  F  cos  <p). 
From   which 

F  sin  <p 


W  +  F  cos  <p' 


(5) 


If  the  weight  of  the  body  be  small  in  comparison  with 
the  applied  force,  then  W  in  equation  (5)  may  be  neg- 
lected, and  we  have 

H  =  tan  <p (6) 

Comparing  equations  (4)  and  (6)  it  is  seen  that  for 
motion  to  be  impending,  the  angle  which  the  applied 
force  makes  with  the  normal  must  be  equal  to  the 
angle  of  repose.  For  any  angle  smaller  than  this, 
there  can  be  no  motion,  no  matter  what  the  intensity 
of  the  applied  force. 

Work  and  Energy.  When  a  force  overcomes  a  re- 
sistance through  space  it  is  said  to  do  work;  and  the 
quantity  of  work  is  measured  by  the  product  of  the 
displacement  of  the  point  of  application  of  the  force 
and  the  component  of  the  force  parallel  to  the  dis- 
placement. In  symbols, 

W  =Fd'} 

where  W  is  the  work,  d  the  displacement,  and  F  the 
component  of  the  applied  force  parallel  to  the  dis- 
placement. 

As  an  example,  assume  as  in  Fig.  9,  the  line  of 
direction  of  the  force  F  making  an  angle  6  with  the 


FORCE  AND   FRICTION — WORK  AND   ENERGY     31 

path   in   which   the   point   of   application  of   the   force 
is  constrained  to  move,  and  suppose  further,  that  the 


d 


A 

FIG.  9. 

point  of  application  A,  is  displaced  through  the  dis- 
tance d,  to  the  point  B. 

The  work  then  is 

W  =  F  cos  6  X  d  =  F  d  cos  d. 

F  cos  6  being  the  component  of  the  force  paral- 
lel to  the  motion. 

Suppose  the  force  applied  to  a  body  which  offers 
no  resistance  excepting  its  own  inertia;  then,  from 
a  previous  formula,  we  have, 

F  =  m  a (a) 

The  distance  passed  over  will  be,  if  the  force  be  con- 
stant so  that  the  acceleration  is  constant,  and  the 
body  start  from  rest, 

2         2  a 

Multiplying  the  equations  (a)  and  (6),  member  by 
member,  we  have 

=  W  -  ^.  (7) 


In  this  case,   work  is  consumed   in   imparting  motion 


32  ELEMENTARY   MECHANICS 

to  the  mass  m;  but  the  mass  in  being  brought  to 
rest  is  able  to  do  work  in  quantity  precisely  equal 
to  that  consumed  in  imparting  the  speed  v.  Work  is 
therefore  said  to  be  stored,  or  the  body  possesses 
Energy, — capability  of  doing  work — in  virtue  of  its 
motion.  The  energy  a  body  possesses  due  to  its 
motion  is  called  Kinetic  Energy. 

When  a  body  is  able  to  do  work  due  to  its 
position,  its  energy  is  said  to  be  potential.  Examples 
of  potential  energy  are  bodies  above  the  earth's  sur- 
face, compressed  springs,  etc. 

Unit  Work.  In  the  c.g.s.  system,  the  unit  work  is 
the  erg,  and  is  equal  to  the  work  done  in  bringing 
abo.ut  a  displacement  of  one  cm.  against  a  force  of 
one  dyne.  The  practical  unit  work  is  the  joule,  and 
is  equal  to  io7  ergs. 

In  the  F.P.S.  system,  the  unit  work  is  the  foot 
poundal,  and  is  equal  to  the  work  done  in  bringing 
about  a  displacement  of  one  foot  against  a  force  of 
one  poundal.  The  engineers'  unit,  the  //.  lb.,  is  the 
work  done  in  raising  the  weight  of  one  pound  one 
foot.  One  ft.  lb.,  therefore,  equals  g  foot  poundals; 
or  approximately,  i  ft.  lb.  =  32.2  foot  poundals. 

Conservation  of  Energy.  The  change  in  kinetic 
energy  that  a  body  undergoes  in  passing  over  a  given 
path  is  equal  to  the  work  done  in  traversing  that 
path.  For,  assume  that  at  the  beginning  of  the  path 
the  velocity  of  the  body  is  v^  then  after  traversing 


FORCE   AND   FRICTION — WORK   AND    ENERGY      33 

the  distance  s,  if  the  acceleration  is  a,  we  have,  by 
a  previous  demonstration,  the  following  relation: 

v2  •-=  v\  +  2  as\        .      .      .      .      (8) 
where    v    is    the    velocity    at    the    end    of    the    path  s. 

Multiplying    both  sides  of  equation   (8)   by  — ,  where 

m  is  the  mass  of  the  body,  we  have 

mv2      m  v2, 

= \-mas.        .      .      .      (o) 

2  2 

The  second  term  of  the  right-hand  member  of  equa- 
tion (9)  is  the  product  of  a  force  and  distance,  and 
therefore  represents  work.  Writing  equation  (9)  in 
another  form,  we  have 


-mas.      .      .      .     (10) 


If  the  acceleration  is  not  constant,  we  may  assume 
the  path  to  be  divided  into  n  elements,  each,  element 
of  the  path  being  indefinitely  small,  such  that  the 
acceleration  for  any  element  may  be  considered  con- 
stant. Rewriting  equation  (10)  for  this  condition,  we 
have 


2  2 

—  (m  al  st  +  m  a2  s2  +  m  a3  s3  +  .  .  .  .  4-  m  an  sn)  ;     .     (i  i) 

where  st,  s2,  s3,  etc.,  represent  the  elements  into  which 
the  path  is  divided,  and  alt  a2,  a3,  etc.,  the  correspond- 
ing accelerations  while  passing  over  those  elements. 
The  left-hand   member  of  equation   (n),  being  the 


34  ELEMENTARY   MECHANICS 

initial  kinetic  energy,  is  a  constant  quantity  for  any 
given  case;  hence,  the  right-hand  member,  which  is 
the  sum  of  the  kinetic  and  potential  energies,  is  a 
constant  quantity  for  any  body  moving  under  the 
action  of  forces  without  collision  with  other  bodies. 
And  this  is  true  for  any  body  of  a  system  of  bodies; 
providing  always  that  there  are  no  collisions,  and  the 
forces  acting  are  due  to  the  mutual  interaction  of  the 
bodies,  and  not  to  actions  external  to  the  system.  It 
follows,  then,  that  the  total  energy  of  such  a  system 
remains  constant. 

Potential  and  kinetic  energy  of  masses  are,  however, 
not  the  only  forms  of  energy.  But  by  suitable  pro- 
cesses, energy  may  be  transformed  from  one  form  into 
another.  One  of  the  simplest  cases  of  transformation 
of  energy  is  the  destruction  of  kinetic  energy  and  the 
simultaneous  evolution  of  heat  through  friction  or 
impact. 

If  in  any  system,  from  which  no  energy  escapes, 
and  into  which  no  energy  enters,  account  be  taken 
of  all  forms  of  energy,  then  no  matter  what  trans- 
formations take  place  within  the  system,  the  sum 
total  is  a  constant  quantity.  This  is  the  principle  of 
the  conservation  of  energy.  It  is,  so  far  as  experience 
goes,  consistent  with  all  physical  phenomena. 


CHAPTER   IV 

MOMENTUM,  PRINCIPLE  OF  MOMENTS, 
AND  IMPACT 

Momentum.  By  momentum  is  understood  the 
quantity  of  motion  a  body  possesses,  and  is  expressed 
numerically  by  the  product  of  mass  and  velocity.  In 

symbols 

M  =  m  v (i) 

If  a  body  whose  mass  is  m,  having  an  initial  veloc- 
ity i>,,  has  applied  a  constant  force  F,  its  velocity 
will  be  augmented  at  a  uniform  rate.  Let  its  velocity 
at  the  end  of  the  time  /  be  represented  by  va;  then 
the  change  in  momentum  is 

M'=  m  (v2  -  v,}\ 
but 

^  m  (v.,  -  vt) 

I 
therefore 

///  =  M  =  m  (v2  -  Vt).      ...     (2) 

Or  the  change  in  momentum  is  equal  to  the  product 
of  force  and  time;    i.e.,  the  Impulse. 

The  Three  Laws  of  Motion 

(i)  Every  body  continues  at  a  uniform  rate  of 
motion  in  a  right  line,  unless  compelled  to  change  its 
rate  by  some  force  external  to  it. 

35 


36  ELEMENTARY   MECHANICS 

(2)  Change    of    momentum    is    proportional    to    the 
impulse  that  produces  it  and  in  the  same  direction. 

(3)  Action  and   reaction  are   equal   and   opposite   in 
direction. 

Statement  (2)  includes  statement  (i) ;  for,  since 
statement  (2)  says  that  change  in  momentum  is  equal 
to  the  impulse  that  produces  it,  it  follows  that  if  the 
impulse  is  zero,  i.e.,  no  external  force,  then  there  is 
no  change  in  momentum,  and  hence  no  change  in 
the  rate  of  motion;  and  further,  it  implies  that  if  the 
impulse  is  zero,  there  is  no  change  in  direction. 

Statement  (3)  merely  declares  that  the  actions  of 
two  bodies  on  each  other  are  always  equal  and  op- 
posite in  direction.  This  action  and  reaction  between 
two  bodies  considered  jointly  is  termed  stress. 

Principle  of  Moments.  In  considering  the  action 
of  a  force  on  a  rigid  body,  it  is  necessary  to  consider 
its  three  elements;  viz.:  Intensity,  Point  of  Application, 
and  Direction. 

It  is  the  result  of  experience  that  the  tendency  of 
a  force,  of  given  intensity,  to  produce  rotation  is 
independent  of  the  point  of  application,  so  long  as 
the  direction  in  which  the  force  acts  remains  un- 
changed. If,  then,  we  understand  by  the  line  of 
direction  of  a  force,  the  direction  and  position  of  the 
line  along  which  the  force  acts,  then  the  conditions  in 
regard  to  the  tendency  of  the  force  to  produce  rota- 
tion are  completely  specified  by  specifying  its  intensity 


MOMENTUM,    AND   PRINCIPLE   OF   MOMENTS     37 

and  its  line  of  direction,  or  as  some  prefer  to  call  it, 
its  line  of  action. 

Moment  of  a  Force.  The  moment  of  a  force,  with 
respect  to  an  axis,  is  a  measure  of  the  tendency  of 
the  force  to  produce  rotation  about  that  axis.  Nu- 
merically, it  is  equal  to  the  product  of  the  force  and 
the  perpendicular  distance  from  the  axis  to  the  line 
of  direction  of  the  force.  Calling  this  perpendicular 
distance  the  arm  of  the  force,  then  the  moment  is 
equal  to  the  product  of  the  force  and  its  arm.  In 
symbols 

C  =  Fd] 

where  C  is  the  moment,  F.  the  force,  and  d  its  arm. 
(As  a  matter  of  convenience,  moments  tending  to 
produce  anti-clockwise  rotation  are  considered  posi- 
tive, and  those  tending  to  produce  clockwise  rotation, 
negative.)  Experiment  shows  that  if  a  rigid  body, 
capable  of  rotating  about  a  fixed  axis,  is  in  equi- 
librium under  the  action  of  any  number  of  forces  lying 
in  a  plane  perpendicular  to  the  axis,  then  the  sum 
of  the  moments,  with  respect  to  that  axis  tending  to 
produce  rotation  in  one  direction,  is  equal  to  the  sum 
of  the  moments  tending  to  produce  rotation  in  an 
opposite  direction;  or,  the  algebraic  sum  of  the  mo- 
ments is  equal  to  zero.  This  theorem  is  called  the 
principle  of  moments. 

The  two  conditions  necessary  so  that  a  rigid  body 
shall  be  in  equilibrium  under  the  action  of  coplanar 


38  ELEMENTARY   MECHANICS 

forces  are:  the  algebraic  sum  of  the  forces  in  any 
direction  must  be  equal  to  zero,  and  the  algebraic  sum 
of  the  moments  about  any  axis,  perpendicular  to  the 
plane  of  the  forces,  must  be  equal  to  zero. 

One  particular  case  requires  especial  attention. 
Assume  a  body  under  the  action  of  a  number  of 
coplanar  forces  acting  in  various  directions,  and  that 
these  forces  be  resolved  into  horizontal  and  vertical 
components.  Assume  further,  that  the  algebraic  sum 
of  these  components  in  any  direction  is  equal  to 
zero;  but  that  the  algebraic  sum  of  the  moments  is 
not  equal  to  zero.  It  is  obvious  that  such  a  system 
of  forces  can  not  be  balanced  by  one  force;  but  that 
two  equal  and  opposite  forces  must  be  applied,  the 
sum  of  whose  moments  is  equal  and  opposite  to  the 
sum  of  the  moments  of  the  system.  Such  a  pair  of 
forces  is  called  a  couple;  that  is,  two  equal  and  op- 
posite forces  not  in  the  same  straight  line,  constitute 

a  couple,    whose    moment    is  equal 
A 

to  one  of  the   forces   multiplied  by 

*""]"'  the  perpendicular  distance  between 

d  their  lines  of  direction.    This   may 

I "  be    shown  as  follows:     Assume,  as 

in  Fig.   10,  the  two   equal  coplanar 
FIG.  10. 

parallel  forces  oppositely  directed, 
with  the  perpendicular  distance  d  between  their  lines 
of  direction.  Choose  any  axis  O  at  a  perpendicular 
distance  x  from  the  line  of  direction  of  the  force  F2. 


MOMENTUM,    AND   PRINCIPLE   OF   MOMENTS       39 

The    moment    of    the    force    Fly    with  respect  to  this 

axis,   is 

.   F,  (d  +  x)  =  F,  d  +  Fl  x. 

The  moment  of  the  force  F2,  with  respect  to  the  same 

axis,  is  _  p2x  =  -  F^x; 

since  Ft  =  F2.  Hence,  the  sum  of  the  moments  of 
the  two  forces,  or  the  moment  of  the  couple  is 

G  =  Fl  d  +  F,  x  -  F!  x  =  Fl  d. 

It  is  clear  that  the  same  result  will  be  obtained  no 
matter  where  the  axis  is  chosen,  and  hence  the 
moment  of  the  couple  is  equal  to  the  product  of  one 
of  the  forces  and  the  perpendicular  distance  between 
their  lines  of  direction. 

Centre  of  Mass.  If  two  particles,  ml  and  w2,  be 
rigidly  connected  and  a  force  applied  on  the  straight 
line  joining  them  at  such  a  point  between  the  two 
particles  that  only  linear  acceleration  is  produced,  the 
moments  about  the  point  of  application,  due  to  the 
reactions  of  the  particles,  must  be 
equal  and  oppositely  directed.  Since  1W 

the   acceleration  is   the  same  for  both      -rr-       ' 

' 
particles,    the  reactions    must    be    di- 

rectly    as  their    masses;     and,    conse- 

FlG.    II. 

quently,     for     the     moments     to     be 

equal,   the   arms   must   be   to   each   other   inversely  as 

the    masses   of   the    particles.     From    which,    we    have 

(Fig.  n), 

ml  dl  =  m2d2, 


40  ELEMENTARY   MECHANICS 

but  for  this  to  be  true,  we  must  have 

yn^  /t  =  m2 12, 

and  the  point  c  is  determined.  The  point  c  is  the 
centre  of  mass  of  the  two  particles,  and  is  so  situated 
that  if  a  force  be  applied  at  this  point,  in  any  direc- 
tion whatsoever,  the  two  particles  will  be  equally 
accelerated  parallel  to  the  direction  of  the  applied 
force;  and,  hence,  no  motion  of  rotation  is  produced 
in  the  system.  If  there  be  a  third  particle,  w3,  then 
to  find  the  centre  of  mass  of  the  three  particles, 
we  assume  a  mass  equal  to  ml  +  m2  concentrated  at 
c,  and  combine  this  with  w3,  in  precisely  the  same 
manner  that  mv  and  m2  were  combined.  In  a  similar 
manner,  the  centre  of  mass  for  any  number  of 
particles  is  found.  The  centre  of  mass  of  a  system 
of  particles  is  then  a  point  where  the  whole  mass  of 
the  system  may  be  considered  concentrated,  and  is 
so  situated  in  the  system  that  if  forces  be  applied 
whose  resultant  passes  through  the  point,  there  will 
be  no  tendency  to  produce  rotation.  From  the  fore- 
going, discussion  it  is  evident  that  the  centre  of  mass 
is  a  point  so  situated  in  the  body,  that  if  a  plane  be 
passed  through  it  in  any  direction  whatsoever,  then 
the  sum  of  the  products,  obtained  by  multiplying  each 
particle  by  its  perpendicular  distance  from  the  plane, 
on  one  side  of  the  plane,  must  be  equal  to  the  sum 
obtained  in  a  similar  manner  on  the  other  side  of  the 
plane;  or  if  we  call  the  products  on  one  side  of 


MOMENTUM,   AND   PRINCIPLE   OF   MOMENTS       41 

the  plane  positive  and  on  the  other  side  of  the 
plane  negative,  their  algebraic  sum  must  be  zero. 

Centre  of  Gravity.  The  force  with  which  a  body 
is  attracted  toward  the  centre  of  the  earth  is  the  re- 
sultant of  the  forces  due  to  the  particles  of  the  body; 
and  since  the  angle  subtended  at  the  centre  of  the 
earth,  by  any  body  of  ordinary  dimensions,  is  practi- 
cally nil,  the  force  with  which  the  body  is  attracted 
is  the  resultant  of  a  system  of  parallel  forces. 

The  sum  of  these  parallel  forces,  or  the  resultant, 
is,  of  course,  the  weight  of  the  body.  The  point 
where  the  entire  weight  of  the  body  may  be  considered 
to  act  is  called  the  centre  of  gravity,  and  is  deter- 
mined by  the  principle  of  moments  in  precisely  the 
same  manner  as  the  centre  of  mass.  It  is  clear  that 
in  all  ordinary  cases  the  centre  of  mass  and  centre  of 
gravity  are  denned  by  the  same  point.  For  bodies  of 

» 

regular  geometric  form  and  homogeneous  in  con- 
struction the  centre  of  mass  coincides  with  the  geo- 
metrical centre.  For  bodies  not  so  constituted  the 
centre  of  mass  or  centre  of  gravity  must  be  found 
experimentally. 

Impact  and  Momentum.  From  the  third  law  of 
motion  it  follows,  that  if  two  bodies  impinge,  the  forces 
acting  on  the  two  bodies  are  equal;  since  action  and 
reaction  are  equal.  Furthermore,  since  the  time  of 
impact  is  the  same  for  both  bodies  it  follows  that  the 
impulse  acting  on  the  one  is  equal  to  the  impulse 


42  ELEMENTARY   MECHANICS 

acting  on  the  other;  therefore,  the  change  in  momen- 
tum in  the  one  body  is  equal  to  the  change  in  mo- 
mentum in  the  other  body. 

Suppose  two  bodies  whose  masses  are  m±  and  m2 
and  having  velocities  ul  and  uz,  impinge  in  such  a 
way  that  the  velocities  are  changed  to  vl  and  vz. 
Then  since,  from  what  has  been  previously  shown, 
the  changes  in  momentum  are  equal,  it  follows  that 

ml  («!  -  v,)  =  m2  (v2  -  «,),    ...      (3) 
from  which 

ml  ul  +  m2  u2  =  ml  v1  +  m2  v2.       .      .      (4) 

That  is,  the  sum  of  the  momenta  before  impact  is 
equal  to  the  sum  of  the  momenta  after  impact.  This 
is  the  first  law  of  impact. 

If  there  is  no  energy  absorbed  during  impact,  the 
bodies  are  said  to  be  perfectly  elastic,  and  the  kinetic 
energy  of  the  system  after  impact  is  equal  to  the 
kinetic  energy  before  impact;  hence 

ml  u\  +  m2  u22  =  ml  v\  +  m2  v*2,   .      .      (5) 
and 

ml  (u\  -  v2t)  =  m2  (v\  -  u\), 

from  which 

ml  (ul  -  v,)  (ul  +  v,)  =  m2  (v2  -  u2)  (v2  +  w2); 
by  equation  (3) 

MI  Oi  -  vO  =  m2  (v2  -  M2), 
hence 


MOMENTUM,   AND   PRINCIPLE   OF   MOMENTS       43 

from  which 

MI  —  U2  =  ^2  —  Vi> 
and 

(6) 


That  is,  the  ratio  of  the  difference  of  velocities 
after  impact  to  the  difference  of  velocities  before  im^ 
pact  is  a  constant,  and  equal  to  unity.  In  the  case 
of  two  bodies  impinging,  which  are  not  perfectly 
elastic,  this  ratio  is  less  than  unity,  and  must  be  der 
termined  by  experiment.  In  any  case,  the  ratio 


is  a  constant  for  any  two  given  bodies  and  is  called 
the  coefficient  of  restitution. 

Conservation  of  Momentum.  Let  the  two  spherical 
bodies  of  masses  m^  and  w2,  approach  each  other  on 
the  right  line  joining  their  centres  with  velocities,  ul 
and  u2,  as  indicated  in  Fig.  12. 

Let  a  b  be  the  plane  of  impact,  and  the  full  circles 
represent  the  position  of  the  bodies  with  respect  to 
the  plane  a  b  one  second  beiore  impact.  Let  the 
centre  of  mass  of  the  two  bodies  at  that  instant  be 
to  the  right  of  the  plane  by  a  distance  x.  The 
centre  of  mass  of  the  body  mly  one  second  before 
impact,  must  be  to  the  right  of  the  plane  of  impact 
by  the  distance  ul  +  r^  the  centre  of  mass  of  the 
body  m2,  must  at  the  same  instant  be  to  the  left  of 


44 


ELEMENTARY   MECHANICS 


the   plane   a  b  by   the   distance   u2  +  r2.     Writing  now 
the  equation  for  the  centre  of  mass,  we  have 

m,  (u,  +  rl-  x)  =  m2  (u2  +  r2  +  x) ; 
from  which 


w,  u,  — 


mlrl  —  m2r2 


m 


m 


(8) 


If   now   the   dotted   figures,    displaced   downward   to 
avoid    confusion,    represent    the    positions    of    the    two 


•W2 


I 

r,  I 


FIG.   12. 

bodies  one  second  after  impact,  and  the  centre  of 
mass  of  the  system  is  at  that  instant  to  the  left  of 
the  plane  a  b  by  the  distance  y,  then  by  writing  the 
equation  for  the  centre  of  mass  for  this  configuration, 
we  obtain 


r,)  =  m2  (v2  +  r2  -  y), 


MOMENTUM,   AND   PRINCIPLE  OF  MOMENTS       45 

from  which 

mlvl  +  m2  v2  ml  rl  -  m^  r2 

-  -   =  y  -f-  -  —  .       .         .        (O) 

-- 


The  numerators  in  the  left-hand  members  of  equa- 
tions (8)  and  (9)  represent  respectively  the  sum  of 
the  momenta  before  and  after  impact,  and  are  there- 
fore equal;  hence  the  right-hand  members  of  these 
two  equations  are  equal.  Now  the  fraction, 

mlrl  —  m2r.2 
mv  +  m2 

represents  the  distance  that  the  centre  of  mass  of  the 
two  bodies  is  to  the  right  of  the  plane  of  impact  at 
the  instant  of  impact.  Hence  in  the  unit  of  time 
before  impact  the  centre  of  mass  passes  over  the 
distance  ^r^  -  m2r2 


and  in  the  unit  of  time  after  impact  the  centre  of 
mass  moved  over  the  distance 

ml  rl  —  m2r2 
ml  +  m2 

But,  by  equations  (8)  and  (9),  these  two  distances 
are  equal,  and  hence  the  rate  of  motion  of  the  centre 
of  mass  is  unchanged  by  the  impact. 

If  the  two  bodies  do  not  approach  each  other  on 
the  same  straight  line,  but  on  lines  inclined  to  each 
other,  then  the  impact  will  be  oblique.  In  this  case 
the  velocities  of  the  bodies  may  each  be  resolved  into 
two  components,  one  at  right  angles  to  the  plane  of 


46  ELEMENTARY  MECHANICS 

impact  and  one  parallel  to  it.  The  components 
parallel  to  the  plane  of  impact  will  be  unchanged, 
and  those  perpendicular  to  the  plane  of  impact  will 
obey  the  laws  of  direct  impact.  Hence,  that  part  of 
the  component  of  the  velocity  of  the  centre  of  mass 
perpendicular  to  the  plane  of  impact  is  the  same  after 
impact  as  before,  and  the  parallel  component  being 
unaffected,  it  follows  that  the  velocity  of  the  centre  of 
mass  is  unchanged. 

It  will  be  readily  seen  how  this  demonstration  can 
be  extended  to  three  or  more  bodies.  This  constitutes 
the  fundamental  law  of  momentum;  viz.:  The  veloc- 
ity of  the  centre  of  mass  of  a  system  cannot  be 
altered  by  any  internal  forces;  or  in  other  words 
the  momentum  of  a  system  can  be  changed  only  by  the 
action  of  a  force,  or  forces,  external  to  that  system 

Loss  of  Energy  during  Impact.  As  previously 
stated,  there  is  no  loss  of  energy  when  perfectly 
elastic  bodies  impinge.  If,  however,  the  bodies  are 
not  perfectly  elastic  then  there  is  a  loss  of  energy 
depending  upon  the  coefficient  of  restitution. 

To  obtain  the  expression  for  loss  of  energy  during 
impact,  it  is  convenient  to  first  obtain  equations  for 
the  final  velocities  in  terms  of  the  masses,  the  initial 
velocities,  and  the  coefficient  of  restitution. 

From   the   principle   that   the   velocity   of  the   centre 
of  mass  is  unchanged  by  impact,  we  may  write 
(MI  +  m2)  V  =  ml  ul  -f  m2  u2  =  ml  vl  +  m2 1>2; 


MOMENTUM,    AND   PRINCIPLE   OF   MOMENTS       47 


where  V  is  the  velocity  of  the  centre  of  mass.      From 
this  we  obtain 

^  ul  4-  m.2  u2 


and 


V  = 


V  = 


ml  +  m2 


m 


ml  +  m2 

From  equations  (7)  and  (n),  we  obtain 

m, 


(10) 

(II) 


m 


e  (MJ  -  «2) 


m, 


(12) 


Substituting  in  (12)   for  V  the  value  as  given  in  (10), 
we  obtain 


•      •      (13) 


^-  w2 

4-  w, 


,  —  w2) 


The  kinetic  energy,  before  impact,  is  equal  to 


-(mlu\ 


and,  after  impact,  it  is  equal  to 
i 

2          1       l 

The  loss  in  kinetic  energy  then  is 


E 


m2  u      — 


(14) 


48  ELEMENTARY   MECHANICS 

Substituting  in  equation  (14),  the  values  of  vl  and  v2 
as  given  in  (13),  performing  the  indicated  operations 
and  simplifying,  we  obtain,  finally 


CHAPTER  V 


MOMENT    OF    INERTIA 

ASSUME  a  force  F  applied,  as  indicated  in  Fig. 
13,  and  that  the  particles  mlt  m2,  m3,  .  .  .  mn,  are 
situated  at  distances  rn  r2,  r3,  .  .  .  rw,  from  the  axis 
O,  and  are  rigidly  connected  to  this 
axis.  Assume  further,  that  all  masses 
may  be  neglected  but  the  masses  of  w3(p<  f 
the  particles,  and  that  there  is  no  , 
friction;  then  the  various  masses  will  , 
be  accelerated,  and  every  particle  ex- 
erts a  reaction  due  to  its  inertia. 

Let  these  reactions  be  represented 
by  /i,  /2,  /»•  •  •  •  fn,  and  the  linear 
accelerations  by  a15  a2,  a.3,  .  .  .  a, 


d 


j£)< fo 


FIG.  13. 
we  then  have 


w 


(I) 


49 


ELEMENTARY   MECHANICS 


and,   from  the  Principle  of  Moments,   we  have 

Fd=firi+f2r2+f3r,+ +fnrn     -     (2) 

The  bodies  being  rigidly  connected,  the  angular  ac- 
celeration must  be  the  same  for  all;  hence,  from 
equation  (i),  we  obtain 

r,  ~  m,  r, 
^L  =     f* 

0_n  =        fn 

rn      WH  rn 
from  which 

yt  =  a  mlr1 
f2  =  a  m2  r2 

Substituting  these  values  of  /1?  /,,  /3,  .  .  .  fn,  in 
equation  (2)  we  obtain 

F  d  =  a  ml  r2:  +  a.  m2  r22  -f  «  m3  r23  + +  OL  mn  r2u. 

a  being  the  same  for  all  terms,   we  have 


=  a  I  m  r2 


(5) 


MOMENT   OP  INERTIA  $1 

It  is  obvious  that  the  preceding  demonstration  holds 
for  any  number  of  masses;  and  that  in  any  case 
where  there  is  a  rigid  body  capable  of  rotating  about 
a  fixed  axis,  we  would  obtain 

Fd  =  G  =  (I  mr2)  a;  .  .  .  (6) 
in  which  G  is  the  torque,  or  turning  moment,  and 
I  m  r2  is  the  summation  of  the  products  obtained  by 
multiplying  the  mass  of  each  particle  by  the  square 
of  its  distance  from  the  fixed  axis. 

It  is  obvious  that  for  any  given  body  with  given 
axis,  I  m  r2  is  a  constant ;  hence,  we  may  replace  it 
by  a  symbol;  i.e., 

G  =  /  « (7) 

As  previously  defined,  G  is  a  measure  of  the  ten- 
dency of  a  couple  to  produce  rotation  about  a  given 
axis;  hence,  it  follows  that  /  has  precisely  the  same 
relation  to  motion  of  rotation  that  mass  has  to  mo- 
tion of  translation.  It  has  been  stated  that  the  in- 
ertia of  a  body  may  be  measured  by  the  force  re- 
quired to  produce  a  given  linear  acceleration.  In 
the  same  way,  /  may  be  measued  by  the  torque, 
or  turning  moment,  required  to  produce  a  given 
angular  acceleration.  Hence,  /  may  be  appropriately 
called  Moment  of  Inertia. 

We  may  then  define  the  moment  of  inertia  of  a 
body  with  respect  to  an  axis  as  a  measure  of  the  im- 
portance of  the  inertia  of  that  body  as  regards  its 
rotation  about  that  axis. 


52  ELEMENTARY   MECHANICS 

It  is  of  such  importance  to  the  engineering  student 
to  have  a  proper  conception  of  moment  of  inertia, 
that  it  is  advisable  to  derive  formula  (7)  by  an  en- 
tirely independent  method;  i.e.,  from  the  principle  of 
energy. 

Assume  again,  as  previously,  that  the  particles 
whose  masses  are  mlt  m2)  m3,  .  .  .  mn)  are  rigidly 
connected  to  the  axis  O,  as  in  Fig.  13,  at  distances 
r\->  r2>  rv  •  •  •  rn-  If  at  any  instant  the  angular 
velocity  of  the  system  about  the  axis  is  w,  then  the 
linear  velocity  of  the  various  particles  is  a>  rly  w  r2,  w  r.3, 
.  .  .  CL>  rn  ;  and  the  kinetic  energy  of  the  system  is 


Ek  =  - 

222 


(8) 


The    angular    velocity    being    the    same    in   all    terms, 
equation  (8)   may  be  simplified  as  follows: 


Ek  =  Y  (ml  r\  +  w2r%  +  m9r\  +  ......  +  mnr\) 

=  ^2mr*',    ........      ...     (9) 

where    I  m  r2    has    precisely  -the    same    significance    it 
had    in   equation    (6). 

Hence,  replacing  it  by  the  symbol  /,  we  obtain 

I  co2 
Ek  =  —  ......      (10) 

As  has  been  previously  shown,  the  kinetic  energy  of 


MOMENT   OF   INERTIA  53 

-a  body  expressed  in  terms  of  its  mass  m,  and  velocity 

v,  is 

mv2 
Ek  =  — . 

Comparing  this  expression  with  equation  (10),  it 
is  at  once  seen  that  /,  the  moment  of  inertia,  has  the 
same  relation  to  motion  of  rotation  that  mass  has  to 
motion  of  translation. 

The  moment  of  inertia  of  all  homogeneous  bodies 
that  have  regular  geometric  figures  may  be  found  by 
computation;  but  the  student  must  take  this  for 
granted  until  he  has  learned  to  integrate.  It  will  be 
shown  later  how  to  find,  by  experiment,  the  moment 
of  inertia  of  any  body  that  is  not  homogeneous  or  of 
regular  geometric  form. 

Change  of  Axes.  If  the  moment  of  inertia  of  a 
body  with  respect  to  a  given  axis  is  known  it  is 
always  possible,  by  a  simple  computation,  to  find  the 
moment  of  inertia  about  a  new  axis,  parallel  to  the 
given  axis,  and  at  a  fixed  distance  from  it. 

Let  the  moment  of  inertia  of  the  body,  as  depicted 
in  Fig.  14,  about  the  axis  A  B,  passing  through  the 
centre  of  mass  C,  be  Ic.  To  find  the  moment  of 
inertia  about  the  axis  E  D,  parallel  to  the  axis  A  B, 
and  at  a  distance  a  from  it.  Consider  any  particle 
of  mass  m,  situated  at  a  distance  r  from  the  centre 
of  mass.  From  the  figure,  we  have 
x2  =  a2  +  r2  +  2  a  b, 


54 
and 


ELEMENTARY   MECHANICS 


m  x2  =  m  a2  +  m  r2  +  m  2  a  b. 

Repeating  this  operation  for  each  particle,  we  have 
I  m  x2  =  I  m  a2  +  2  m  r2  +  I  m  2  a  b.    .    (n) 

But,  a  being  a  constant,  we  may  write  equation   (n) 
as  follows: 

2  m  x2  =  M  a2  4-  I  m  r2  4-  2  a  I  m  b]    .     (12) 

where    M  is  the   mass  of  the  body.      Now   I  m  x2  is 
the  moment  of  inertia  about  the  axis  E  D,  and  may 


FIG.  14. 

be  represented  by  the  symbol  I0.  I  m  r2  is  the 
moment  of  inertia  of  the  body  about  the  axis  A  B\ 
i.e.,  about  the  axis  passing  through  the  centre  of 
mass  C.  I  m  b  represents  the  moment  of  all  the 
particles  about  the  axis  C  and  is  by  a  previous 
proposition  equal  to  zero. 

Hence,  equation  (12)  may  be  written, 

I0  =  Ic  +  Ma2 (13) 

Therefore,  the  moment  of  inertia  with  respect  to  any 


MOMENT   OF    INERTIA 


55 


axis,  parallel  to  an  axis  passing  through  the  centre  of 
mass,  is  equal  to  the  moment  of  inertia  with  respect 
to  the  axis  through  the  centre  of  mass,  plus  the  mass 
multiplied  by  the  square  of  the  distance  to  the  par- 
allel axis. 

This  may  also  be  proven  by  the  principle  of  energy. 
Assume    the    body    whose    mass    is    M    and    centre    of 
mass  at  C,  Fig.   15,  to  be  revolving  with  a  uniform 
angular  velocity  &>,  about  the 
axis    through    O,    in    such   a 
manner  that  any  line  on  the 
body,  such  as  b  d,  is  always 
parallel   to   its  first   position. 
Its  kinetic  energy  then  is 


(14) 


of VX—L 


FIG.   15. 


If,    however,    the    body    is 

rigidly  connected  and  rotates  about  the  axis  through 
O,  in  such  a  manner  that  the  line  b  d  makes  a  con- 
stant angle  0  with  the  radius  0,  as  depicted  in  the 
second  position  by  V  d',  the  body  must  rotate  about 
the  axis  through  C  with  an  angular  velocity  equal  to 
the  angular  velocity  of  the  radius  a  about  the  axis 
O.  The  kinetic  energy  of  the  body,  due  to  its  rota- 
tion about  the  axis  through  C,  is 


(15) 


56  ELEMENTARY   MECHANICS 

Adding  equations  (14)  and  (15)  we  obtain,  for  the 
total  kinetic  energy 

W~^(Ic  +  Ma*).       .      .      .     (16) 

As  has  been  previously  shown  the  kinetic  energy  of  a 
rotating  body  is  equal  to  the  product  of  the  moment 
of  inertia,  and  one  half  the  square  of  the  angular 
velocity.  Hence,  the  moment  of  inertia  of  the  body 
about  the  axis  through  O,  is  equal  to  the  moment  of 
inertia  about  a  parallel  axis  through  the  centre  of 
mass,  plus  the  mass  multiplied  by  the  square  of  the 
distance  between  the  two  axes,  as  has  been  shown  by 
equation  (13). 

There  are  certain  cases  where  it  is  possible,  by  a 
simple  method,  to  find  the  moment  of  inertia  of  a 
figure  about  an  axis  which  is  not  parallel  to  the  axis 
about  which  the  moment  of  inertia  is  known. 

In  Fig.  1  6,  let  ABC  be  a  triangle  of  mass  M. 
Then  the  moment  of  inertia  of  the  triangle,  about 

the  axis  XX,  is 

Mh* 


where  h  is  the  altitude  of  the  triangle. 

If  the  three  sides  of  the  triangle  are  given,  or  two 
sides  and  the  included  angle,  or  two  angles  and  the 
included  side,  then  the  whole  triangle  is  determinate, 
and  hence  it  is  possible  to  determine  the  area,  and 
the  mass  m,  per  unit  area. 


MOMENT   OF   INERTIA  57 

Let  it  be  required  to  find  the  moment  of  inertia  of 
the  triangle  about  a  new  axis  Xl  Xly  lying  in  the 
same  plane  and  passing  through  the  point  A,  and 
making  an  angle  6  with  the  axis  X  X. 

The  triangle  ABC  being  determinate,  the  altitude 
and  area  of  the  triangles  A  B  D  and  AC  D  can  be 
determined. 

Assume  that  the  mass  per  unit  area  of  the  triangle 
AC  D  is  the  same  as  that  of  the  triangle  ABC. 


FIG.   16. 


Then  by  a  simple  computation  the  mass  of  the  tri- 
angle A  B  D  may  be  determined,  and  its  altitude 
being  known,  its  moment  of  inertia  with  reference  to 
XlXl  may  be  stated;  i.e., 


where    Ml    is    the    mass    and    h^    the    altitude    of    the 
triangle   A  B  D.      In    a   like    manner    the    moment    of 


$8  ELEMENTARY  MECHANICS 

inertia  of  the  triangle  AC  D  with  respect  to  the  axis 
X,X,  is 

,       M2  h\ 
~6^; 

where  M2  is  the  mass  and  h2  the  altitude  of  the 
triangle  AC  D.  Now  the  moment  of  inertia  of  the 
triangle  ABC  with  respect  to  the  axis  X^  Xl  is  equal 
to  the  moment  of  inertia  of  the  triangle  A  B  D,  minus 
the  moment  of  inertia  of  the  triangle  AC  D  with 
respect  to  the  same  axis.  In  symbols: 

7-7        /     -  M^2i       M*h2*  (  R. 

'3  -  A  -  '2  -     —         —£—  •       •       (18) 

Again,  since  the  moment  of  inertia  about  any  axis 
is  equal  to  the  moment  of  inertia  about  the  axis 
passing  through  the  centre  of  mass,  plus  the  product 
of  the  mass  and  the  square  of  the  distance  to  the 
new  axis  parallel  to  the  given  axis,  we  have 

7c  =  /3-Mtf2;  .  .  .  .  (19) 
where  a  is  the  distance  between  Xl  X^  and  X2  X2) 
and  Ic  is  the  moment  of  inertia  about  the  axis  X2X2 
passing  through  the  centre  of  mass.  Let  d  be  the 
distance  between  the  axis  X2  X2  and  some  parallel 
axis  X3  X3.  Then,  by  the  same  principle,  the  moment 
of  inertia  with  respect  to  X3  X3  is 

74  =  Ic  +  M  d2  =  /3  -  M  a2  +  M  d\     .     (20) 

Since  6  may  be  any  angle,  it  follows  that  the  moment 
of  inertia  of  the  triangle  ABC  may  be  found  with 


MOMENT   OF   INERTIA  59 

respect  to  any  axis  lying  in  the  plane  of  the  figure. 
Furthermore,  it  is  obvious  that  the  foregoing  demon- 
stration applies  to  any  plane  figure  that  is  determinate, 
whose  moment  of  inertia  about  an  axis  lying  in  the 
plane  of  that  figure  is  determinate. 

Radius  of  Gyration.  Since,  as  has  been  previously 
stated,  for  any  given  body  with  given  axis,  the  sum- 
mation I  m  r2  is  a  constant,  we  may  write 

/  =  I  mr2  =  MK2] 

where  M  is  the  mass  of  the  body,  and  K  is  the  dis- 
tance from  the  fixed  axis  where  the  mass  M  would 
have  to  be  concentrated  at  a  point  so  as  to  have  a 
moment  of  inertia  with  respect  to  the  axis  equal  to 
that  of  the  body.  K  is  called  the  radius  of  gyration 
of  the  body,  with  respect  to  the  given  axis,  and  nec- 
essarily changes  in  value  for  different  axes. 


CHAPTER   VI 

POWER  AND  ANGULAR  MOMENTUM 

THE  rate  at  which  work  is  being  done  is  called 
Power.  If  the  point  of  application  of  a  force  is  dis- 
placed through  a  distance  s,  and  the  component  of 
the  force  in  the  direction  of  the  displacement  is  F, 
the  work  done  is  F  s.  If  the  motion  is  uniform,  and 
t  is  the  time  consumed  during  the  displacement,  then 
the  rate  of  doing  work,  that  is,  the  power,  is  constant. 

In  symbols 

W       F  ? 


From  which  we  see   that   the   power   expended   is   nu- 
merically equal  to  the  product  of  the  component  of  the 
force  in  the  direction  of  the  motion  and  the  speed. 
In  like  manner,  it  can  be  shown  that  the  power  ex- 
pended   is    numerically    equal    to    the 
product   of    Torque    and    Angular  Vc- 
locity. 

Let,  as  in  Fig.   17,  the  drum  whose 
FIG.  17. 

centre    is   O,    and    radius    r  be    acted 

upon  by  a  constant  force  F,  and  let  the  resistance 
be  such  that  the  angular  velocity  maintained  is  a 
constant.  '  The  work  done  in  a  time  /  is 

W  =  F  co  r  t, 
60 


POWER  AND  ANGULAR  MOMENTUM  6 1 

and  the  power  is 

W 

P  =  —  =  Fr  w  =  G  to.      .      .      .      (2) 

Again,  let  a  constant  force  F,  act  upon  a  mass  m, 
which  is  perfectly  free  to  move,  then  the  acceleration 

is 

F 

a  =  —  =  a  constant, 
m 

Since  velocity  is  equal  to  the  product  of  accelera- 
tion and  time,  it  follows  that  it  is  a  variable  and 
changes  at  a  uniform  rate  with  the  time;  in  other 
words,  it  is  a  function  of  the  time.  This  being  the 
case,  that  is,  the  force  being  constant,  and  the 
velocity  variable,  it  follows  that  the  power  expended 
to  produce  the  accelerated  motion  is  variable.  But 
power  is  equal  to  the  product  of  velocity  and  force; 
hence,  at  any  instant  the  value  of  the  power  ex- 
pended is  equal  to  the  product  of  the  force  and  the 
instantaneous  velocity;  i.e., 

p  =Fv; 

where  p  represents  instantaneous  value  of  power  and 
v  represents  instantaneous  value  of  velocity. 

In  the  French  system,  the  unit  power  is  doing  work 
at  the  rate  of  one  joule  per  second.  This  unit  is 
called  the  Watt.  In  the  English  system,  the  unit 
power  is  the  Horse-power  and  is  equal  to  doing  work 
at  the  rate  of  33,000  ft.  Ibs.  per  minute. 


62  ELEMENTARY  MECHANICS 

Angular  Momentum  or  Moment  of  Momentum.  If 
a  body  of  mass  m,  has  a  velocity  in  the  circumference 
of  a  circle  whose  radius  is  r,  its  momentum  is 

mv  =  m  oj  r  ; 
where  CD  is  the  angular  velocity. 

The  angular  momentum  then  is 

m  v  r  =  m  co  r2  .....      (3) 

If  some  constant  force  has  been  acting  to  produce 
the  velocity  v,  there  has  been  a  constant  acceleration, 
and  the  force  is 


from  which  we  have  for  torque 

v          mcor2 
G  =  m  -  r  --     —  —  .....      (4) 

Under  the  conditions  stated,  the  velocity  v,  and  the 
angular  velocity  M,  will  be  functions  of  the  time;  i.e., 
will  change  at  a  uniform  rate.  The  power,  however, 
at  any  instant  that  is  expended  to  produce  the  acceler- 

ation -    =  —  ,  is  the  product  of  the  torque  and  angu- 
/         t 

lar  velocity;  i.e., 

mcor2 
p  =  Gco  =     —j—  aj  .....      (5) 

m  CD  r2  . 
But,  ---  is  the  rate  of  change   of   angular  momen- 

tum; therefore,  the  instantaneous  value  of  power  is 
equal  to  the  product  of  rate  of  change  of  angular 
momentum  and  the  instantaneous  angular  velocity. 


CHAPTER   VII 

TENSION    IN    CORDS 

IF  a  cord  support  a  weight,  the  tension  in  the 
cord  at  any  section  is,  of  course,  equal  to  the  weight 
supported  plus  the  weight  of  the  cord  below  the  sec- 
tion. In  practical  problems,  it  is,  however,  seldom 
necessary  to  consider  the  weight  of  the  cord  since  it 
is  usually  a  very  small  fractional  part  of  the  total 
weight  supported. 

As  an  example,  a  Manilla  rope,  capable  of  support- 
ing 7,000  Ibs.,  weighs  YT,  Ib.  per  running  foot.  Assume 
a  length  of  100  ft.,  supporting  a  weight  of  7,000  Ibs.; 
the  weight  of  the  rope  then  is  about  33.3  Ibs.,  which  is  a 
trifle  less  than  %  per  cent  of  the  total  weight  supported. 
Since  the  ultimate  breaking  strength  of  any  specimen 
of  material  can  never  be  predicted  with  certainty 
closer  than  i  per  cent  or  2  per  cent,  it  follows  that 
in  most  cases  the  weight  of  the  cord,  cable,  or  rope 
may  be  neglected. 

As  previously  stated,  if  a  rope  support  a  weight, 
the  tension  in  the  rope  is  equal  to  the  weight.  If, 
however,  the  body  supported  has  an  accelerated 
motion,  the  tension  may  have  any  value. 

To   fix   the   attention,   assume   a   body   supported   as 

63 


64  ELEMENTARY  MECHANICS 

in  Fig.  1 8.     If  the  body  be  at  rest  the  tension  in  the 

cord  is 

T  =  Mg  =  W (i) 

If  the   body   be  raised   or  lowered   with  a   constant 
speed,  the  tension   is  still  M  g;    for,  since  there  is  no 
acceleration    there    can    be    no    additional 
force.      Assume   now   that   the   drum   has 
impressed   upon   it    an    acceleration,    such 
that   the   body  has  an  acceleration  down- 
.  .   .       ward  of  g  feet   per  sec.  per  sec.     The  ten- 
sion,   then,    is     zero;      for    the     body     is 
perfectly  free  to  fall. 

In  general,  if  the  acceleration  downward  is  a,  the 
tension  is  the  unbalanced  force;  i.e., 

T  =  M  (g  -  a) (a) 

If  the  acceleration  is  upward,  it  is  considered  nega- 
tive, and  therefore,  for  upward  acceleration 

T  =  M(g  +  a} (3) 

If  the  downward  acceleration  is  greater  than  g,  it 
follows  from  equation  (2)  that  T  is  negative;  that 
is,  a  pressure.  This  is  possible  only  if  the  support 
is  rigid,  like  a  rod,  and  has  impressed  upon  it  a 
downward  acceleration  in  excess  of  g. 

Take  a  concrete  case  by  assuming  a  spring  bal- 
ance suspended  from  the  roof  of  an  elevator  cage, 
which  is  ascending  with  a  constant  acceleration  of 
1 6  ft.  per  sec.  per  sec.  Assume  further,  that  this 


TENSION   IN   CORDS 


spring  balance  is  supporting  a  mass  of  200  pounds, 
to  find  the  weight  registered  by  it.  From  formula 
(3)  we  have 

T  —  200  (32  -I-  1  6)  poundals; 

g  being  assumed  equal  to  32  ft.  per  sec.  per  sec. 
From  this  we  find 

T  =  9,600  poundals 
=  300  Ibs. 

If,  on  the  other  hand,  the  cage  is  descending  with 
a  constant  acceleration  of  16  ft.  per  sec.  per  sec.,  then 
the  weight  registered  becomes 

T  =  200  (32  —  1  6)   poundals 
=  3,200  poundals 
=  100  Ibs. 

Atwood's    Machine.     Assume,    as    in    Fig.    19,    two 
masses  M  and  m,  supported  by  a  cord,  whose  mass 
is  negligible,  over  a  wheel,  without  mass,  and  perfectly 
free  to  rotate.     To  determine  the  tension 
in  the  cord  and  the  acceleration  of  the 
system.        The     total     mass     moved     is 
(M  +  m)'    and  the  moving  force  is 

F  =  Mg-mg=?(M-m)g.    .    (4) 

The    acceleration    is,    the    moving    force 
divided  by  tlie  mass  moved,  i.e., 

M  —  m 


fim 


FIG.  19. 


s 


(5) 


66  ELEMENTARY  MECHANICS 

The  tension  in  the  cord,  supporting  the  mass  M,  is 

M  -  m 


T  =  M  (g  -  a)  =  A 

\  u        M  +  m  "  y 

-irSi*- (6) 

The  tension  in  the  cord,  supporting  the  mass  m,  is 

M  -  m     \ 


*  U  •  M  +  * 

2  M  m 


M  +  m* 

thus  making  T  and  Tl  equal.  This  must  necessarily 
be  so;  since  the  wheel  is  assumed  without  mass  and 
friction,  the  tension  must  be  the  same  throughout  the 
cord. 

Assume  now,  the  cord  supporting  the  mass  M 
wrapped  over  a  cylinder  of  radius  r,  and  moment  of 
inertia  /,  to  find  the  tension  in  the  cord  and  the 
acceleration. 

If  T  is  the  tension  in  the  cord,  then,  as  has  been 
previously  shown 

Tr  =  Ia, (7) 

T  r  being  the  torque,  /  the  moment  of  inertia,  and 
a-  the  angular  acceleration.  From  which  the  linear 
acceleration  is 

a~ar=-~ (8) 


TENSION  IN   CORDS  67 

As  previously  shown 

T-M(g-a)  =M(g-  •?-£)• 

From  which  we  obtain 


Substituting  in  equation  (8),  the  value  for  T  as  found 
in  equation   (9),  we  obtain 

Mr2 


Dividing  by  r,  we  obtain 


a  Mr 


If  the  drum  is  a  solid  homogeneous  cylinder,  of  mass 

Mly  then 

Mtr2 

2 

Substituting    this    value    in    equations    (9),    (10),    and 
(n),  we  obtain 

MM,  - 
T  =  -  — - s.  =  *      jr.  (12) 


M  r2  +  M,  j 

M  f 2  2  M 

ffl    =    -  o    £    =    TT 


2 

Mr  2 


68 


ELEMENTARY   MECHANICS 


Assume,  now,  a  drum  of  moment  of  inertia  /  and 
radius  r,  having  wrapped  around  it  a  cord  supporting 
a  mass  M  .on  one  side,  and  a  mass  m  on  the  other 
^ — ->^  side,  as  depicted  in  Fig.  20.  Assuming 
no  friction,  we  have 


llm 

rl" 

FIG.  20. 


T  =  —  +  m  (s  +  «) ;    •    (J5) 

and  since  a  =  a  r,  we  have 

T  =  ~  +  m(g  +  ar).    .     (16) 


Again 

T  =  M(g-ar);       . 

hence,  equating  (16)  and  (17),  we  have 

~   +  m  (g  +  a  r)  =  M  (g  -  a  r) ; 

from  which 

la  -f  m  r  g  +  m  a  r2  =  M  r  g  -  Mar*. 
Rearranging  and  factoring,  we  have 

a  ( I  +  m  r2  +  M  r2)  =  (M  -  m)  r  g, 
from  which 


(M  -  m)  r 

a  =  ~ — r— ,  g,    . 


finally, 


a  =  a  r  = 


I  +  (M  +  m)  r2 
(M  -  m)  r2 


(18) 


/  +  (M  +  m)  r2  * 

Substituting  now,  in  equation  (17),  for  a  its  value,  we 
obtain 


TENSION   IN   CORDS  69 


I  +  2 

"M*/+(M 
Also 


I  +  2Mr2 

2' 


In  this  case  there  must  be  a  difference  in  the  tensions, 
since  the  drum  has  inertia;    this  difference  is 

/  +  2  m  r2 


l  =  T  -  t  =  M  g  j 


I  +  (M  +  m)  r2 

I  +  2  M  r  2  I  (M  -  m) 

n  g  I  +  (M  +  m)  r2  ~  I  +  (M  +  m)  r2  *'  ' 

This  may  be  found  directly,  since  the  difference  in 
the  tensions  must  be  equal  to  the  tension  required  to 
produce  the  acceleration  of  the  drum,  i.e., 

la        I  (M  —  m)  r 

1=:~     =       X      +  (M  +  m)  r2  g 


I  (M  -  m) 


£> 


I  +  (M  +  m) 

which  is  the  same  as  given  in  equation  (22). 

Reverting  to  the  figure,  we  see  that  the  moving 
force  is  (M  —  m)  g,  and  the  mass  moved  is,  M 
+  m  +  K;  where  K  is  the  equivalent  mass  of  the  drum. 
The  acceleration  then  is 

M  —  m  ,    . 

a--~g-      '      '      '      (23) 


70  ELEMENTARY   MECHANICS 

As  expressed  in  equation  (19) 

(M  -  m)  r2 
=  7  +  (M  +  m)  r2  g' 

Hence,  by  equating  (19)  and  (23),  we  have 

i_  _  r2 

M  +  m  +  K  =  =  r  +  (M  +  m)  r29 
and 

7  +  M  r2  +  m  r2  -  M  r2  +  w  r2  +  K  r2; 

from  which 


The  value  of  K  may  also  be  found  in  a  different 
manner.  Let  a  constant  force  F  be  applied  to  the 
surface  of  a  drum  whose  moment  of  inertia  is  7,  and 
radius  r,  then 

F  r  =  7  a, 

and 


from  which 
But 

7~: 
Fr2 

7 

where  K  is  the  equivalent  mass  of  the  drum,  hence, 

~K  "=  Fr 


from  which 


K  =  —,  as  before. 
r2 


CHAPTER   VIII 

MAXIMA    AND    MINIMA 

IT  frequently  becomes  necessary,  when  dealing  with 
equations  involving  two  or  more  variables,  to  deter- 
mine under  what  condition  one  of  these  variables 
attains  either  a  maximum  or  a  minimum  value  in 
terms  of  the  other  quantities  involved.  In  general, 
this  is  done  by  the  application  of  the  differential 
calculus.  It,  however,  is  frequently  the  case  that  the 
conditions  are  such  as  to  enable  us  to  do  this  by 
inspection,  or  by  elementary  mathematics.  A  few 
examples  will  be  given  here  to  illustrate  this. 

Let  it  be  required  to  determine  the  value  of  the 
angle  which  makes  the  product  of  its  sine  and  cosine 
a  maximum.  Given  the  equation 

y  =  sin  x  cos  x]      .      .      .      .      (i) 

to  determine  under  what  conditions  y  assumes  its 
maximum  value.  Equation  (i)  may  be  written  in 
the  form  of 

y  =   —  sin  2  x (2) 

Since  the  sine  has  its  maximum  values-  when  the 
angle  is  90°,  450°,  etc.,  and  its  minimum  values  when 
the  angle  is  270°,  630°,  etc.,  we  have  y  In  equation 

71 


72  ELEMENTARY   MECHANICS 

(2),  and  consequently,  the  product  of  the  sine  and 
cosine  of  x  a  maximum,  when  x  equals  45°,  225°, 
etc.;  and  the  minimum  values  occur  for  the  product 
of  sine  and  cosine  when  x  equals  135°,  315°,  etc. 

Let  it  be  required  to  show  what  must  be  the  rela- 
tion between  two  variable  quantities,  whose  sum  is  a 
constant,  such  that  their  product  is  a  maximum.  Let 

where  k  is  a  constant  and  x  and  y  variables.  Make 
k,  as  in  Fig.  21,  the  diameter  of  a  circle,  inscribe  the 
triangle  b  c  d,  and  drop  a  perpendicular  a,  from  the 
point  c,  dividing  the  diameter  k  into 
the  segments  x  and  y.  Since  b  c  d  is 
a  right-angled  triangle,  we  have 

a2  =  x  y.     .  .      (4) 

FIG.  21.         If,   now,    the    vertex    c    take    various 

positions    along    the    circumference    of 

the  circle,  k   remains   constant,    but   x,  y,   and   a  will 

vary;  and  a  attains  its  maximum  value  when  equal  to 

k 

-.       But   when  this   occurs,   x   and  y  are    equal  and 

k 

each  equal  to  — .  Hence,  from  equation  (4),  it  fol- 
lows that  the  product  of  two  variables,  whose  sum  is 
constant,  is  a  maximum  when  the  two  variables  are 
equal. 

We  may  now  draw  the  further  conclusion  that  the 
triangle  of  maximum  area  which  can  be  inscribed  in 


MAXIMA  AND  MINIMA  73 

a  semicircle  is  the  one  having  equal  legs.  For  the 
area,  which  is  measured  by  the  product  of  the  diam- 
eter k,  and  altitude  a,  is  a  maximum  when  a  is  a 
maximum,  but  this  occurs  when  the  legs  b  c  and  c  d 
are  equal.  Stated  more  broadly,  the  area  of  any 
right-angled  triangle  of  given  hypothenuse  and  variable 
legs  is  a  maximum  when  the  legs  are  equal. 

It  will  now  be  shown  that  if  a  right-angled  triangle 
have  the  perpendicular  distance  from  the  vertex  of 
the  right  angle  to  the  hypothenuse  a  constant,  and 
the  legs  variable,  the  hypothenuse  is  a  minimum  when 
the  legs  are  equal.  Let,  in  Fig.  22,  the  triangle  b  c  d 
have  the  legs  b  c  and  c  d 
equal,  and  an  altitude  equal 
to  a,  making  the  hypothenuse 
equal  to  2  a.  Let  the  tri- 
angle b  e  f  have  the  legs  b  e 
and  ef  not  equal,  then  the 

angle  6  is  less  than  45°.  Calling  the  segments  x 
and  y,  then  the  hypothenuse  is  equal  to 

a  a 

= 


tanO       tan  (90  -  6)  ' 
from  which 


x  +  y 


sin  0  cos 


but   it   has  been  shown  that   the  product  of  the  sine 
and    cosine    is    a    maximum    when    the    angle    is    45°. 


74  ELEMENTARY  MECHANICS 

Hence,  since  6  is  less  than  45°,  this  product  is  less 
than  — ,  and  we  have  x  +  y  >  2  a.  The  hypothenuse 

is  therefore  a  minimum  when  the  legs  or,  what 
amounts  to  the  same  thing,  the  two  segments  are 
equal. 

We  may  now  draw  the  further  conclusion,  since 
x  y  =  a2,  that  when  the  product  of  two  variables  is  a 
constant,  their  sum  is  a  minimum  when  the  variables 
are  equal. 

Given  the  particle  a,  Fig.  23,  moving  in  the  direc- 
tion a  h  with  the  constant  speed  u  and  the  particle 
b  moving  in  the  direction  b  h  with  the  constant  speed 
vt  together  with  the  distance  a  b  and  the  angles  6 


and  <p,  to  determine  what  will  be  the  position  of  the 
particles  when  the  distance  between  them  is  a  mini- 
mum. The  relative  motion  of  the  two  particles  is 
precisely  the  same  as  though  b  were  at  rest  and  a 
had  impressed  upon  it,  together  with  its  own*  velocity, 
a  velocity  a  d,  parallel  and  equal  to  v  but  in  the 


MAXIMA  AND  MINIMA  75 

opposite  direction.  The  resultant  velocity  of  a,  under 
these  conditions,  is  a  i,  and  since  b  is  at  rest  the 
particles  are  nearest  together  when  the  particle  a  is 
at  the  point  g;  the  point  g  being  found  by  dropping 
a  perpendicular  from  b  on  to  the  line  a  I  produced. 
Actually  at  that  instant  the  particle  a  is  at  a';  the 
point  a'  being  on  the  intersection  of  the  line  a  h 
produced  and  a  line  through  g  parallel  to  b  h.  And 
the  particle  b  will  be  at  b',  the  point  b'  being  on  the 
intersection  of  the  line  b  h  and  a  line  through  a' 
parallel  to  g  b.  The  angles  6  and  <p  being  given, 
the  angle  /?  is  determinate,  and  hence  the  magnitude 
and  direction  of  a  i  may  be  found.  Knowing  the 
direction  of  a  b  and  a  g,  the  angle  a  is  given;  this, 
together  with  a  b,  determines  a  g  and  b  g,  and  from 
these  a  a'  and  b  b'  may  be  found. 

Given  the  two  particles  a  and  b,  Fig.  24,  a  having 
a  constant  speed  v,  in  the  direction  a  c  making  an 
angle  6  with  the  line  a  b',  to  determine  what  is  the 
minimum  speed  which  b  may  have  and  still  meet  the 
particle  a.  Let  x  be  the  speed 
with  which  the  particle  b  is  mov- 
ing along  some  line  b  c-}  the  path 

b  c  making    an    angle    <p   with  the 

FIG.  24. 
line    a  6,  and    the    point    c    being 

the  intersection  of  their  paths.  If  t  is  the  time 
required  for  the  particle  a  to  travel  from  a  to 
c,  then  a  c  =  v  t  ;  and  for  the  particles  to  meet 


76 


ELEMENTARY   MECHANICS 


at  c,  we  have   b  c  =  x  t.      From  the  law  of  sines  we 
have  xt  \v  t:\sin  0  :  sin  < 


from  which 


x  =  v 


sin  0 
sin  <p' 


For  x  to  be  a  minimum,  sin  <p  must  be  a  maximum, 

since  v  and   sin  6  are   constant.       But   the   maximum 

value   for   sin  <f>    is   unity,   and   occurs   when   <p  equals 

90°;    hence,  the  minimum  value  for  the  speed  of  b  is 

x  =  v  sin  6. 

Having  given  the  coefficient  of  friction  between  a 
body  and  an  inclined  plane  together  with  the  angle 
of  inclination  of  the  plane,  to  determine  the  minimum 
force  that  will  move  the  body,  without  acceleration, 
up  the  plane. 

Let,  in  Fig.  25,  W  represent  the  weight  of  the 
body,  then  the  normal  pressure  on  the  plane,  due  to 

the  weight  W,  is 

R,  =  W  cos  6- 

where  0  is  the  angle  of  inclination  of  the  plane.     If, 

now,  F  represents  the  re- 
quired force  making  an  angle 
(p  with  the  plane,  its  normal 
component  is 

R2  =  F  sin  <p, 


FIG.  25. 


and    the    normal    pressure   on 
the  plane  is  the  algebraic  sum  of  Rv  and  R.2;    i.e., 

R  =  W  cos  6  -  F  sin  <p.     .      ,      .  (5) 


MAXIMA   AND   MINIMA  77 

The  component  of  the  force  F,  parallel  to  the  plane 
must,  to  maintain  a  constant  speed  up  the  plane,  be 
equal  to  the  component  of  the  weight  parallel  to  the 
plane  plus  the  force  of  friction.  The  force  of  friction 
being  equal  to  the  product  of  the  normal  pressure 
and  coefficient  of  friction,  we  have 

Fcostp  =  WsinO  +  p  R;  .  .  .  (6) 
where  /*  is  the  coefficient  of  friction.  Substituting 
in  equation  (6)  the  value  of  R  as  given  in  equation 
(5),  we  obtain 

F  cos  (p  =  W  sin  6  +  /*W  cos  6  —  p.  F  sin  <p\ 
from  which,  solving  for  F,  we  find 

sin  6  +     cos  6 


cos  <p  +  IJL  sn  <p 

The  coefficient  of  friction  is  equal  to  the  tangent  of 
the  angle  of  repose.  If  we  then  represent,  by  ft,  the 
angle  of  repose,  we  have 

sin  ft 

**  =  ^ri' 

and  substituting  this  value  of  //,   in  equation   (7),  we 
obtain 


F  = 


sin  6  H  ---  cos  6 


sin  ft    . 

cos  <p  4-  -     -  sin  <p 
cos  ft 


This  expression  reduces  to 


m         ... 

cos  (<p  -  ft) 


78  ELEMENTARY   MECHANICS 

From  equation  (8)  it  is  evident,  since  the  numerator 
is  a  constant,  that  F  is  a  minimum  when  <p  and  ft 
are  equal,  making  the  denominator  a  maximum. 

To  determine  the  pitch  of  the  thread  of  a  jack- 
screw,  having  given  the  coefficient  of  friction,  such 
that  its  efficiency  shall  be  a  maximum.  The  thread 
of  the  screw  being  an  inclined  plane  and  the  applied 
force  acting  parallel  to  the  base  of  the  plane,  we 
will  first  determine  this  force  in  terms  of  the  weight 
lifted  and  the  coefficient  of  friction. 

Let,  in  Fig.  26,  F  be  the  force  required  to  move 
the  weight  W  at  a  uniform  rate  up  the  plane,  and  <p 
the  angle  of  inclination  of  the  plane.  The  total 

normal    pressure    on    the    plane, 

due    to    the    weight    W    and    the 

force  F,  is 

R=Rl+R2=W  cos<p  +  Fsin<p.    (9) 

The   force   component   parallel   to 
the  plane,  being  the  sum   of   the 

weight  component  parallel  to  the  plane  and  the  force 
of  friction,  we  have 

f  =  W  sin<p  +  pR.  .  .  .  (10) 
Substituting  in  equation  (10),  the  value  of  R  as  given 
in  equation  (9),  we  obtain 

f  =  W  sin  ^  +  jj.W  cos  <p  +  pF  sin  <p. 
Dividing  both  sides  of  the  equation  by  cos  y>,  we  have 
F  =  W  tan  <f>  +  p  w  +  V  F  tan  <P\ 


MAXIMA   AND   MINIMA  79 

and  solving  for  F,  we  find 


. 

i  —  ft.  tan  <f> 

Substituting   in  equation    (n),   for    //,  the  value  -  -; 
where  p  is  the  angle  of  repose,  we  obtain  ) 


sin  <p       sin  p 


p  =  w   cos 


sin  p  sin  <p' 
cos  p  cos  <p 
from  which 

F  =  W  Sm  ^  +  ^.  (12) 

cos  (p  +  (p) 

If  now  d  is  the  diameter  of  the  screw,  the  height 
through  which  the  weight  is  lifted,  during  one  revolu- 
tion, is 

h  =  TT  d  tan  <p\ 

and  since  the  work  done  by  the  screw  is  measured 
by  the  product  of  the  weight  and  height,  we  have 

The  work  done  on  the  screw,  during  one  revolution, 
is 

TT7  77         J  TJ/  Sin   (P    +    ^)          J  /        \ 

W«  =  F  7t  d  =  W 7 ^f  TT  d.    .     .     (14) 

cos  (P  +  <?} 

The  efficiency  of  any  machine  being  the  ratio  of  the 
work  done  by  the  machine  to  the  work  done  on  the 


8o  ELEMENTARY   MECHANICS 

machine,  we  find,  by  dividing  equation  (13)  by  equa- 
tion (14), 

W  r  d  tan  <p 


cos  (p  +  <f>) 
from  which 

=  sin  <p  cos  (p  +  tp) 
V  ~  sin  (3  +  <p)  cos  <p 


Since,    sin  x  cos  y  =  —  sin  (x  +  y)   +    -  sin  (x  --  y), 

equation   (15)    may  be   written  in  the  form  of 

sin  (p  +  2  (p)  —  sin  p 
~  sin  (p  +  2  <p)  +  sinp' 
from  which 

_      __  2  sin  p  _  . 

77  "  sin  (p  +  20+  sin  p' 

TI  is  a  maximum,  when  the  second  term  of  the  right- 
hand  member  of  equation  (16)  is  a  minimum;  and 
this  occurs,  since  the  numerator  is  a  constant,  when 
the  denominator  is  a  maximum;  and  since  p  is  a 
constant,  the  denominator  is  a  maximum  when  p 
+  2  <p  =  90°.  Hence  for  77  to  be  a  maximum 

v  =  45°  -  £ 

Having  given,  the  direction  in  which  a  vessel  is  to 
sail,  the  direction  of  the  wind,  and  furthermore,  as- 
suming no  drift  and  the  sail  a  plane  surface,  to  de- 
termine the  set  of  the  sail  such,  that  the  component 


MAXIMA  AND  MINIMA  8 1 

of   the    wind   pressure    producing    motion,  shall    be    a 
maximum. 

Let,  in  Fig.  27,  c  d  be  the  direction  in  which  the 
vessel  is  to  sail,  and  P  represent  in  magnitude  and 
direction  the  pressure  of  the  wind  on  the  sail  a  b. 
Resolving  P  into  two  com- 
ponents, the  component 


parallel  to  the  sail  has  no   & Xff —   -^ — 

effect    in    producing    mo-  p    \     '  ' 


tion;  and  the  normal  com- 
ponent is  FlG-  27' 

fb  =  P  sind (17) 

Now,  resolving  fb  into  two  components,  one  normal 
to  the  direction  of  the  motion,  which  produces  drift 
and  is  here  neglected,  and  the  other,  parallel  to  the 
direction  of  the  motion,  and  which  produces  the  mo- 
tion, is 

F=fbsin<p (18) 

Combining  equations  (17)  and  (18),  we  obtain 

F  =  P  sin  6  sin  <p (19) 

Now,  since  the  direction  of  the  wind  is  fixed  and  also 
that  of  the  motion  of  the  vessel,  we  have  the  angle 
/?  a  constant,  and  further,  since 

<f>  +  6  =  /?, (20) 

we  may  eliminate  one  of  the  variable  angles  in  equa- 
tion (19).      Substituting  in  equation  (19),  the  value  of 
<p  as  found  from  equation  (20),  we  obtain 
F  =  P  sin  6  sin  (/?  -  6) ; 


82  ELEMENTARY  MECHANICS 

from  which 

F  =  P  (sin  6  cos  6  sin  p  -  sin2  6  cos  /?).     .     (21) 
Since 


sin  6  cos  6  sin  p  =  —  sin  2  6  sin  /?, 
2 


and  also, 


-  sin2  6  cos  p  =  —  (cos  2  6  cos  p  —  cos  /?) ; 
2 

equation  (21)  reduces  to 

p 

F  =  —  (sin  2  6  sin  p  -f  cos  2  6  cos  p  —  cos  p) ; 

from  which 

F  =  -  [cos  (2  6  -  ft)  -  cos  p] (22) 

Since  cos  /?  is  a  constant,  the  expression  in  the  brace, 
and   hence   F,   becomes   a    maximum    when    2  6  =  p-t 

and  6  =  <f>  =  — ,  which  determines  the  set  of  the  sail 
for  the  maximum  force  in  the  direction  of  motion. 


CHAPTER  IX 

PENDULAR    MOTION 

ANY  body  free  to  vibrate  about  a  fixed  axis  under 
the  action  of  gravity  is  called  a  pendulum. 

The  period  of  a  pendulum,  or  time  of  vibration, 
represented  by  T,  is  the  time  required  to  pass  through 
a  cycle,  i.e.,  it  is  the  time  that  elapses  between  any 
two  successive  identical  positions  when  the  body  is 
moving  in  the  same  direction. 

Half  a  period,  or  the  time  of  an  oscillation,  repre- 
sented by  t,  is  the  time  required  to  pass  through  half 
a  cycle. 

The   amplitude   of  the   pendulum   is   the   maximum 
displacement    from    the    position    of 
equilibrium. 

Simple  Pendulum.  Assume,  as 
depicted  in  Fig.  28,  a  small  particle 
of  mass  m,  concentrated  at  a  point, 
supported  by  a  weightless  cord 
whose  length  is  L.  The  force 
acting  vertically  is  constant  and 
equal  to  m  g,  and  may  be  repre- 
sented by  the  line  a  c.  Resolving  this  into  two 

components,   a  b,   parallel   to   the   motion,    and    b  c    at 

83 


84  ELEMENTARY   MECHANICS 

right  angles  to  the  motion,  or  parallel  to  the  sup- 
porting cord,  then  the  component  producing  motion  is 

a  b  =  F  =  m  g  sin  6  .....      (i) 

Since  acceleration  equals  force  divided  by  mass,  we 
have,  for  the  acceleration  along  the  arc 

a  =  g  sin  6  ......      (2) 

If  the  angular  displacement  of  the  body  be  small,  the 
angle  and  sine  are  sensibly  equal,  and  we  have 

<*  =  g  0  ......      (3) 

But  the  displacement  of  the  body  from  the  position 
of  equilibrium,  measured  along  the  path  described 
by  the  body,  is  proportional  to  the  angular  displace- 
ment; hence,  the  acceleration  is  proportional  to  the 
displacement,  and  the  body  has  a  simple  harmonic 
motion.  Therefore 

a  =  gd  ==  y^-s;  (4) 

where  5  is  the  displacement.     But, 

s  =  LO-, 
and,  substituting  this  value  in  equation   (4),  we  have 


and 


from  which 

(5) 


PENDULAR   MOTION 


Such  an  arrangement  as  we  have  just  been  con- 
sidering is  called  a  simple  pendulum.  It  is,  however, 
impossible  to  realize  this  condition  practically;  since 
any  support  that  may  be  used  has  weight,  and  the 
supported  mass  is  always  a  body  of  finite  dimensions. 

Physical  Pendulum.  Assume  a  rigid  body,  as  in 
Fig.  29,  whose  centre  of  mass  is  at  C,  supported  by 
an  axis  S,  perpendicular  to  the 
plane  of  the  paper.  Let  the  mo- 
ment of  inertia  of  the  body, 
whose  mass  is  M,  about  the  axis 
5  be  /,  then 

M  g  d  =  la]     .      .      (6) 
where  M  g  d  is  the  torque.      Now, 

d  =  r  sin  6 ; 

where  6  is  the  angular  displace- 
ment from  the  position  of  equili- 
brium, and  r  the  distance  from 
the  axis  of  suspension  to  the  centre  of  mass.  Hence, 

M  g  r  sin  0  =  I  a (7) 

If   the   angular   displacement   be   small,   the   sine   and 
angle  are  sensibly  equal,  and  we  may  write: 

M  gr6  =  I  a; (8) 


FIG.  29. 


from  which 


a  = 


Mgr 


(9) 


The  body  being  rigid,  the  angular  acceleration  for  all 


86  ELEMENTARY   MECHANICS 

points  is  the  same  at  any  instant,  and  varies  directly 
as  the  angular  displacement  6.  But  since  angular 
acceleration  and  angular  displacement  are  to  each 
other  directly  as  linear  acceleration  and  linear  dis- 
placement, it  follows  that  every  point  in  the  body  has 
a  simple  harmonic  motion;  therefore, 


Mgr26       47T2 
a  =  a  r  =       --          -      s.       .      .     (10) 


But, 
hence, 

from  which 


4  »' 


, 
M  gr' 

and 


=    2  7T  -     — .       •         •         (l  l) 

gr  ^     g 

Comparing  equation   (u)    with  equation   (5),   we  find 

that    —  -   takes    the   place   of  L:    hence   v^   is    the 
Mr  Mr 

length  of  the  equivalent  simple  pendulum. 

Such  an  arrangement  as  just  discussed  is  called  a 
physical  pendulum.  The  quantity  M  r  is  called  the 
statical  moment;  and  the  length  of  the  physical  pen- 
dulum then  is  numerically  equal  to  the  moment  of 
inertia  about  the  axis  of  suspension  divided  by  the 
statical  moment. 


PENDULAR  MOTION  87 

Kater's  Pendulum.  The  most  accurate  method  for 
determining  the  acceleration  of  gravity  is  by  means  of 
a  pendulum.  But  the  only  quantity  in  equation  (n) 
that  is  readily  determined  by  experiment  is  the 
time. 

It  is,  however,  possible  by  employing  a  Kater's,  or 
reversible  pendulum,  to  determine  the  length,  without 
knowing  the  moment  of  inertia  or  statical  moment. 
The  following  discussion  will  make  this  clear. 

In  Fig.   30,   let   a  b  be   a   rigid   rod   sup-          " 
porting    the    two    unequal    masses   ml  and      <T_>wi 
m2,   and   let   M  be   the   mass  of  the  whole      _ 
system  whose  centre  of  mass    is  at  C.     Let 
this  system  be  supported  by  an  axis  5,  then       j 
the  length  of  the  equivalent  simple  pendu-     _± 
lum  is 

L  =  JTr>      '      '      •     (I2)          » 

FIG.  30. 
where  I8  is   the   moment   of   inertia   of  the 

system   about    the   axis   5,  and    r   the    distance   from 
the  axis  S  to  the  centre  of  mass. 
But,  as  has  been  previously  shown, 

IS  =  IC  +  Mr*; 

where   Ic  is  the   moment  of   inertia  about   a    parallel 
axis  through  the  centre  of  mass.     Hence,  we  have 


*T 


1 


88  ELEMENTARY   MECHANICS 

If  the  pendulum  be  now  reversed  and  suspended 
by  the  axis  O,  parallel  to  the  axis  S,  and  at  a  dis- 
tance L  from  it,  we  will  have  for  the  new  length 

I,  +  M(L-  rY 


'  M(L-r)      •  •      ' 

From  equation  (13),  we  have 

Ic  =  M  r  L  -  M  r2  =  M  r  (L  -  r). 

Substituting   this   value,    for   Ic   in    equation    (14),    we 

obtain 

M  r  (L  -  r)  +  M  (L  -  r)2 

M  (L  -r) 
from  which 

Li=r  +  L-r  =  L.       .      .      .      (15) 

Showing  that  the  length  is  the  same,  and  therefore 
the  time  of  vibration  is  the  same  when  vibrating  about 
the  axis  S  as  it  is  when  vibrating  about  the  axis  O, 
at  a  distance  L  from  S. 

If,  therefore,  we  take  such  an  arrangement,  as 
depicted  in  the  figure,  and  adjust  the  axes  S  and 
O,  until  the  time  of  vibration  about  the  two  axes  is 
the  same,  it  becomes  necessary  only  to  note  the  time, 
and  measure  the  distance  L;  then  by  equation  (5) 
g  may  be  calculated. 

Ballistic  Pendulum.  Assume  a  body,  Fig.  31,  such 
as  a  b  of  mass  M,  and  centre  of  mass  at  C.  If 
a  force  F  be  applied  to  the  body  at  a  distance  x 
from  the  centre  of  mass  C,  the  effect  of  this  force 


PENDULAR   MOTION  89 

will    be    to   produce    a    linear    acceleration;     which    is 


b 


Further,  there  will  be  an  angular  accelera- 
tion about  the   centre  of  mass;    which  is 


Fx 


where     Ic     is     the     moment     of     inertia 
FlG'  3It     about    the    axis    through  C.      The    con- 
dition  for    the  point    S    to  be    at   rest  is 

a  =  a  r\ 

where  r  is  the  distance  between  the  points  S  and  C. 
Substituting  for  a  and  a,  as  given  above,  we  have 

_F  _  Fxr 
M~~      Ic    ' 

from  which 


But  L,  the  distance  from  S  to  the  point  of  application 
of  the  force  F,  is  (x  +  r);    therefore, 

Ie  lc+Mr> 

-JTr^  ~W7~ 

But  (Ic  +  M  r2)  is  the  moment  of  inertia  of  the  body 
about  the  axis  through  S,  hence 

L  =  Wr 


9o 


ELEMENTARY   MECHANICS 


Showing  that  the  distance  from  the  axis  of  suspen- 
sion, where  a  blow  must  be  struck,  so  that  there  shall 
be  no  jar  on  the  axis,  is  the  length  of  the  equivalent 
simple  pendulum;  hence,  the  axis  of  oscillation  is 
also  the  axis  of  percussion. 

Graphical  Representation  of  the  Pendulum.  Con- 
sider any  irregular  body,  such  as 
is  depicted  in  Fig.  32,  having  its 
centre  of  mass  at  C,  and  suspended 

by  an  axis  through  S,  perpendicu- 
\ 
\   lar  to  the  plane  of  the  paper.     If 

/  its  moment  of  inertia  about  an  axis, 
parallel  to  the  axis  through  S,  and 
passing  through  the  centre  of  mass 
C  is  Ic,  and  the  distance  from  the 
centre  of  mass  to  an  axis  of  sus- 
pension is  a,  then  by  a  previous 
equation  the  length  of  the  pendulum  is 


FIG.  32. 


Ma 

where  M  is  the  mass  of  the  body. 

If  now,  with  C  as  centre  and  a  as  a  radius  we 
describe  a  circle,  then  the  axis  of  suspension  may  be 
taken  anywhere  on  the  circumference  of  this  circle 
for  a  constant  time  of  vibration;  for,  the  expression 
for  the  length  of  the  pendulum  is  obviously  constant. 

Let,  now,  O  be  the  axis  of  oscillation,  then  if  we 


PENDULAR  MOTION  91 

describe  a  circle  with  C  as  centre  and  b  as  radius,  b 
being  the  distance  from  C  to  O,  the  time  of  vibration 
of  the  body  when  suspended  at  any  point  on  the  cir- 
cumference of  the  circle  whose  radius  is  b  will  be 
constant  and  the  same  as  when  suspended  on  the 
circumference  of  the  circle  whose  radius  is  a]  and 
the  length  of  the  pendulum  is 

L  =  a  +  b  ......     (16) 

Also 

Ma2      I   +  Mb2  , 


T 


M~o 


Taking,  now,  the  general  equation  for  the  length  of 
the  pendulum  and  writing  it  in  a  different  form,  we 
have 


since  Ic  and  M  are  constant  for  the  body  under  con- 
sideration, then  if  a  be  varied  and  approach  infinity 
for  its  value,  L  approaches  infinity  for  its  value;  and, 
if  a  approach  zero  for  its  value,  L  again  approaches 
infinity  for  its  value.  But,  for  any  other  values  of 
a,  L  will  have  a  finite  value.  We  will  now  show 
that  L  is  a  minimum  when  a  =  b.  Let 

Ic  =  MK2-, 

where  K  is  the  radius  of  gyration  with  respect  to  the 
centre  of  mass,  and  is  that  distance  from  the  centre 
of  mass  where  the  mass  of  the  body  would  have  to 


ELEMENTARY   MECHANICS 


be  concentrated  at  a  point  so  as  to  have  a  moment 
of  inertia  Ic.     In  general 

K=- -\~jtf (J9) 

Let  in  the  triangle  O  S  A,  Fig.  33,  S  O  be  the  length 
of  the  pendulum,  a  and  b,  respectively,  the  distances 
from  the  centre  of  mass  to  the  axis  of 
suspension  and  oscillation.  Erect  a  per- 
pendicular at  C,  and  make  it  equal  in 
length  to  K,  the  radius  of  gyration  for 
the  body  about  an  axis  through  the 
centre  of  mass  C.  We  then  have 

>IG.  33.  IS=I,  +  Ma2  =  MK2 

+  Ma2  =  M  (a2  +  K2),      .      (20) 
and 

I0  =  Ic  +  Mb2  =  MK2  +  Mb2  =  M  (b2  +  K2);  .  (21) 
where  I8  and  I01  respectively,  are  the  moments  of 
inertia  with  respect  to  the  axis  through  S  and  O. 

By  construction,  we  have 

a2  +  K2  =  p2, 
and 

hence,   by  substituting  in  equations  (20)   and   (21),  we 
obtain 

I,  =  MP\ (22) 

and 

I0  =  Mf (23) 

From  equations  (22)    and  (23),    it   follows  that    p  and 
q,  respectively,  are  the  values  for  the  radius  of  gyra- 


PENDULAR   MOTION  93 

tion  for  the  body  when  suspended  by  the  axis  through 
S,    and    when  suspended    by  the  axis  through  O. 

Since  L,  which  is  equal  to  (a  +  b),  is  the  same 
whether  the  body  be  suspended  by  an  axis  through 
S  or  O,  we  have 

M  tf 


=  a  +  b, 

=  a  +  b. 


Ma 
and 

M  q2 
Mb 
Therefore 

p2  =  a2  +  a  b, 
and 

q2  =  a  b  +  62; 
from  which 

f  +  f  --  (a  +  b)\  .  .      .     (24) 

Equation  (24)  shows  that  the  angle  SAO  is  a  right 
angle.  Now,  a  and  b  are  variables  and  K  is  a  con- 
stant, and  as  has  been  previously  shown,  the  hypothe- 
nuse  of  a  variable  right  triangle,  when  the  perpendicular 
distance  from  the  vertex  of  the  right  angle  to  the 
hypothenuse  is  fixed,  is  a  minimum  when  the  hypothe- 
nuse  is  divided  equally  and  is  double  the  perpendicular. 
Therefore,  the  minimum  length  of  the  pendulum  is 

Lm  =  2K (25) 

Equivalent  Mass  of  the  Pendulum.  The  equivalent 
mass  of  the  pendulum  must  be  a  mass  of  such  value 
that  if  concentrated  at  the  point  O,  its  moment  of 
inertia  with  respect  to  the  axis  through  S  is  the  same 


94  ELEMENTARY   MECHANICS 

as  that  of  the  body  under  consideration;  and  when 
reversed  the  mass  concentrated  at  S  must  have  a 
value  such  that  its  moment  of  inertia  with  respect  to 
an  axis  through  O,  is  the  same  as  that  of  the  body; 
and  further,  their  relation  must  be  such  that  their 
centre  of  mass  falls  at  C. 

Let  m0  represent  the  mass  to  be  concentrated  at  O, 
and  ma  the  mass  to  be  concentrated  at  S',    then 

m0  (a  +  b)2  =  M  (a2  +  K2)  =  M  (a2  +  ab), 
and 

Ma 

m°  =  a  +  b (26) 

Again 

ms  (a  +  b)2  =  M  (b2  +  K2)  =  M  (b2  +  a  b), 
and 

Mb 

^-JTTl (27) 

From  which,  by  adding  equations  (26)   and  (27), 

m0  +  m,  =  M  (—^-7  +  -    —, ,}  =  M. 

\a  +  b       a  +  b/ 

To  have  the  same  centre  of  mass  the  moments  about 
C  must  be  equal;  i.e.,  msa  should  be  equal  to  m0  b. 
Multiplying  equation  (26)  by  b  and  equation  (27)  by 
a,  we  find  the  two  expressions  equal.  Therefore  the 
statical  and  dynamical  conditions  are  completely  ex- 
pressed by  assuming  two  masses;  M  -  —  situated 

at  O,  and  M  -        -  situated  at  S. 
a  +  b 


PENDULAR   MOTION 


95 


Conical  Pendulum.  Assume,  as  in  Fig.  34,  a  mass 
m  supported  from  an  axis  O,  and  caused  to  rotate 
such  that  the  supporting  cord  L  describes  a  cone. 
The  mass  then  moves  in  the  cir- 
cumference of  a  circle  of  radius 
r,  and  the  horizontal  force  acting 

m  v2  i 

upon  the  mass   is  -    — ;    where  v          ' 


FIG.  34. 


is  the  speed  in  the  circumfer- 
ence of  the  circle  described  by 
the  mass  m. 

The  vertical  force  is  m  g,   and 
the    condition    of     equilibrium    is 
given  by  the  fact  that  the  direction  of  the  supporting 
cord   prolonged   is  the   diagonal  of  the    parallelogram 

/yyj    tii" 

constructed    upon   the    forces       —    and  m  g  as  sides; 
from  the  similarity  of  triangles 

m  v2 


:  m  g  :  :  r  :  h. 


From  which 


h 


m  v 


(28) 


Let,  now,  n  be  the  number  of  revolutions  the  mass 
makes  per  unit  time  in  the  circumference  of  the 
circle,  then 

v  =  2  n  r  n, 
and 

v2  =  4  7T2  r2  n2. 


96  ELEMENTARY   MECHANICS 

Substituting    this    value    of    v,    in    equation    (28),    we 
obtain 

h  =  —fir  I  =  —  V^  -     (29) 

4X2r2ri*  47T2rc2' 

or,  writing  this  in  another  form,  we  obtain 


This    is    the    equation    for    the    conical    or    centrifugal 


Since  the  time  of  a  revolution  is  T  =  — ,   it  follows 

n 

that 


If,  now,    r   be   small,    so   that    h   and   L   are  sensibly 
equal,  equation  (31)  becomes 


T    =  2  n  ^  — (32) 

Showing  that  the  period  of  a  conical  pendulum  of 
small  amplitude  is  equal  to  that  of  a  simple  pendu- 
lum of  small  amplitude. 


CHAPTER   X 

FALLING    BODIES    AND    PROJECTILES 

SINCE  the  acceleration  of  gravity  is  sensibly  a  con- 
stant for  ordinary  heights  above  the  earth's  surface  at 
any  specified  place,  it  follows  that  the  formulae  de- 
duced in  Chapter  I,  for  uniformly  varied  motion 
apply  equally  for  bodies  moving  under  the  action  of 
gravity;  it  only  becomes  necessary  to  replace  a  by 
g;  where  g  is  the  acceleration  due  to  gravity. 

Making  these  substitutions,  we  obtain: 

"i  =  v0  +  g  t,      .....      (i) 


k  =  Vot  +  *-+-;        ....     (3) 

v0  being  the   initial,   and  v^  the  final  velocities,  t  the 
time,  and  h  the  height. 

If  the  initial  velocity  be  zero,  these  formuke  become, 
v-gt,    .      .      .  (4) 

*-r, a 

(6. 


The   relation  between   velocity   and   height   is   given 
by  formula   (5) ;  i.e.,   to  produce  a  velocity  v  a  body 


97 


98  ELEMENTARY   MECHANICS 


must  fall  from  a  height  h  such  that  v  =  V2  g  h;  and 
similarly  to  rise  to  a  height  h,  the  body  must  be  pro- 
jected with  a  velocity  v  such  that  the  same  relation 
subsists.  The  time  is  fixed  by  either  equation  (4) 
or  equation  (6),  depending  upon  whether  v  or  h  is 
given. 

If  a  body  be  projected  horizontally,  its  range  de- 
pends upon  its  initial  velocity,  and  height  above  the 
earth's  surface;  whereas,  if  it  be  projected  vertically 
the  height  to  which  it  will  rise  depends  solely  upon 
the  initial  velocity. 

If  the  body  be  projected  so  that  its  initial  velocity 
is  inclined  to  the  horizontal  its  height  and  range  both 
depend  upon  the  magnitude  and  direction  of  the 
initial  velocity. 

If  V  is  the  initial  velocity  and  6  the  angle  of  in- 
clination, then  the  horizontal  and  vertical  components 
are  given  as  follows: 

u  =  V  cos  6 
v  =  V  sin  0 

where  u  is  the  horizontal,  and  v  the  vertical  com- 
ponent. 

The  time  required  by  the  body  to  reach  its  highest 

v 
point   is   /  =  — ;    and    since,    in    falling,   an   equal   in- 

o 

terval  is  consumed,  the  total  time,  or  time  of  flight  is 
2V  __  2  V  sin  0 

J-     =    '  — (o) 

g         g 


(7) 


FALLING   BODIES   AND   PROJECTILES  99 

Since  the  horizontal  component  of  the  velocity  is 
constant,  the  range  is  numerically  equal  to  the  pro- 
duct of  the  time  and  horizontal  component.  Desig- 
nating the  range  by  R,  we  have 

2V          2  F2  sin  0  cos  6 

*  =  TW  = ~g '•    '    '   (9) 

This  is  a  maximum  for  a  given  speed,  when  the 
angle  of  inclination  is  45°;  since  the  product  of  the 
sine  and  the  cosine  of  an  angle  is  a  maximum  when 
the  angle  is  45°. 

The  actual  velocity  at  any  instant  is  numerically 
equal  to  the  square  root  of  the  sum  of  the  squares  of 
the  horizontal  and  vertical  velocities  at  that  instant. 

Designating  this  by  vt,  we  have 

vt  =  V  u2  +  (v  -  g  t)>,  .  .  .  (10) 
which  is  a  minimum,  and  equal  to  u  when  at  the 
highest  point ;  since  at  that  instant  (v  —  g  t)  =  o. 

To  obtain  the  equation  for  the  path  of  the  projec- 
tile, we  let  oc  equal  the  horizontal  distance,  and  y  the 
vertical  distance;  we  then  have 

x  =  ut, (n) 

and 

y  =  vt  -g-^-.        .      .      .      .     (12) 

Substituting  in  equation  (12),  the  value  of  t  as  found 
in  equation  (n),  we  obtain 

u2  y  =  uv  x  -  -  x2\  ,      .      .      .      (13) 

which  is  the  equation  of  a  parabola. 


CHAPTER   XI 

ELASTICITY 

FORCE  has  been  defined  as  that  which  changes  or 
tends  to  change  the  rate  of  motion  of  a  body.  But 
since  to  every  action  there  is  an  equal  and  contrary 
reaction,  it  follows  that  there  can  never  be  a  single 
force. 

The  mutual  interaction  of  bodies  changing  or  tend- 
ing to  change  their  rates  of  motion  is  called  a  stress; 
or,  in  other  words,  force  is  a  stress  considered  in  one 
of  its  aspects. 

Heretofore,  we  have  been  considering  bodies  as  be- 
ing perfectly  rigid.  This  is  never  the  case.  When- 
ever a  body  is  under  the  action  of  a  stress,  there 
is  produced  a  change  in  dimensions;  which  may  be 
a  change  in  volume,  a  change  in  shape,  or  as  is 
usually  the  case,  a  change  in  volume  and  shape.  This 
change  is  called  a  strain. 

It  is  the  result  of  experiment,  known  as  Hooke's 
Law,  that  when  a  body  serves  to  transmit  a  stress, 
then  up  to  a  certain  limit,  the  strain  produced  is 
proportional  to  the  applied  stress;  beyond  this  limit, 
the  strain  increases  at  a  greater  rate  than  the  applied 
stress. 

The   force  of  restitution,   or  the  resistance  which  a 


ELASTICITY  IOI 

body  offers  to  a  stress  producing  deformation,  is 
ascribed  to  its  elasticity.  Bodies  which  recover  their 
original  form  upon  the  removal  of  the  applied  stress 
are  said  to  be  perfectly  elastic.  If,  however,  a  body 
be  deformed  beyond  the  limit  for  which  Hooke's  Law 
holds,  it  will  not  return  to  its  original  size  and  shape. 
That  point  where  a  body  ceases  to  obey  Hooke's  Law 
is  called  the  elastic  limit. 

The  ratio  of  the  applied  stress  to  the  corresponding 
strain  in  a  unit  of  a  body  is  numerically  equal  to  its 
modulus  of  elasticity. 

There  may  be  specified: 

(1)  Elasticity  of  traction. 

(2)  Elasticity  of  torsion. 

(3)  Elasticity  of  flexure. 

(4)  Elasticity  of  volume. 

Modulus  of  Tractional  Elasticity.  If  a  body  of 
cross  section  A,  and  length  L  is  subjected  to  a  stress 
S  tending  to  compress  or  elongate  it,  then,  up  to 
the  elastic  limit,  it  is  found  that  the  elongation  e,  is 
directly  proportional  to  the  product  of  the  applied 
stress  and  length,  and  inversely  proportional  to  the 
cross  section.  In  symbols 

SL 

e-— ; 

and 

S  L 

fj.  =  - —    =  a  constant ;     ....     (i) 
A  e 


102  ELEMENTARY   MECHANICS 

where  p.  is  the  modulus  of  tractional  elasticity;  and 
may  be  defined  as  the  ratio  of  the  stress  per  unit 
area  to  the  corresponding  strain  per  unit  length. 

Elasticity  of  Torsion.  Theory  indicates  and  experi- 
ment verifies  that  when  a  cylindrical  body  of  radius 
r  and  length  L  be  clamped  at  one  end,  and  the  other 
end  be  subjected  to  a  couple  G  whose  axis  is  the 
axis  of  the  cylinder,  then  the  amount  of  twist,  or 
torsion  6,  is  proportional  to  the  product  of  the  couple 
and  the  length,  and  inversely  proportional  to  the 
fourth  power  of  the  radius.  In  symbols 

GL 
'"7T- 

The  exact  relation,  between  the  various  magnitudes,  is 
given  by  the  formula 

2GL 

6  =  ^7*> W 

where  n  is  the  modulus  of  rigidity.  Writing  this  in 
another  form,  we  have 

2GL 

n==e^ <3) 

The  modulus  of  rigidity  n,  may  be  determined  in 
two  ways,  one  is  by  direct  measurement;  i.e.,  by 
subjecting  a  cylindrical  body  of  known  length  and 
radius  to  a  given  torque  and  measuring  the  amount 
of  torsion.  These  values  substituted  in  equation  (3) 
determine  n.  By  taking  a  number  of  observations 
and  plotting  a  curve,  torques  as  abscissae  and 


ELASTICITY  103 

amounts  of  torsions  as  ordinates,  the  limit  of  elas- 
ticity may  be  determined  by  noting  the  point  where 
the  curve  departs  from  a  straight  line. 

The  modulus  of  rigidity  may  also  be  determined 
by  clamping  at  one  end  a  cylindrical  body  of  known 
length  and  radius,  and  suspending  from  it  a  mass 
whose  moment  of  inertia  is  determinate,  and  deter- 
mining the  period  of  the  suspended  body  when  vibrat- 
ing about  the  axis  of  the  cylinder. 

Let  T  equal  the  moment  of  torsion;  i.e.,  the  moment 
of  the  couple  which  will  twist  the  body  through  one 
radian.  Then,  since  the  amount  of  torsion  is  pro- 
portional to  the  torque,  it  follows  that 

Or  =  £  =  /«; (4) 

where  /  is  the  moment  of  inertia  of  the  suspended 
body,  and  a  its  angular  acceleration.  This  being 
the  case  it  follows  that  the  torque  tending  to  restore 
the  vibrating  body  to  equilibrium  is  directly  propor- 
tional to  the  angular  displacement.  And,  since  angular 
displacement  and  angular  acceleration  are  directly 
proportional  to  linear  displacement  and  linear  accel- 
eration, it  follows  that  every  point  in  the  body  has 
a  simple  harmonic  motion.  From  this,  it  follows  that 

*-*£••     .....     (5) 
But,  from  equation  (4),  we  have 

r-'.-      ....      .(6) 


104  ELEMENTARY   MECHANICS 

Substituting  in  equation    (6),   the   value  of  a  as  given 
in  equation   (5),  we  obtain 

y:f L  (7) 

Now 


Substituting  this  value  of  G,   in  equation   (3),   we  ob- 
tain 

S-^-     •      •  (8) 


If,  in  place  of  71,  the  time  of  vibration,  we  use  /,  the 
time  of  an  oscillation,  equation  (8)  becomes 

27tIL 

U  =   ^^ k> 

From  equation  (7),  it  is  seen  that  if  the  moment  of 
torsion  of  a  given  wire  be  known,  then  the  moment 
of  inertia  of  the  vibrating  body  is  determined,  no 
matter  how  irregular,  providing  the  time  of  vibration 
is  found.  It  is,  however,  not  necessary  to  know  the 
moment  of  torsion  of  the  wire,  provided  we  first  de- 
termine the  time  of  vibration  of  the  body  whose  mo- 
ment of  inertia  is  sought,  and  then  joining  with  this 
a  body  whose  moment  of  inertia  is  known  (or  can  be 
computed  from  its  dimensions  and  mass),  and  again 
determining  the  time  of  vibration. 

Let  Ix  be  the  moment  of  inertia  of    the  first  body, 


ELASTICITY  105 

and   7\  its  time  of  vibration;    then,   by  equation    (7), 
we  have 


If,  now,  we  join  with  the  body  whose  moment  of 
inertia  is  Ix  a  body  of  moment  of  inertia  /,  and  find 
the  time  of  vibration  T2;  then,  since  the  moment  of 
torsion  is  a  constant,  it  follows  that 


(II) 


•*•     2 

Combining  equations   (10)    and    (n),   we  obtain 


from  which 


Elasticity    of    Flexure.      Assume,    as    in    Fig.    35,    a 
rectangular    beam,    of    width    b    and    depth    2  d,    bent 


35. 


into  the  arc  of  a  circle.  Then  the  innermost  fibres 
will  be  compressed,  and  the  outermost  fibres  will  be 
elongated;  and,  if  the  material  offers  the  same  resist- 
ance to  compression  that  it  does  to  elongation,  the 


106  ELEMENTARY   MECHANICS 

amount  of  compression  of  the  innermost  fibres  will 
be  equal  to  the  amount  of  elongation  of  the  outer- 
most fibres;  and  furthermore,  the  amount  of  com- 
pression at  any  section  at  a  distance  x  from  the 
concave  surface  will  be  equal  to  the  amount  of  elonga- 
tion at  a  distance  x  from  the  convex  surface;  and  at 
a  distance  d  there  will  be  neither  compression  nor 
elongation.  A  plane  passed  through  the  beam  mid- 
way between,  and  parallel  to  the  two  surfaces,  will 
not  change  in  length  when  the  beam  is  bent.  This 
plane  is  called  the  neutral  plane.  If  the  resistance 
offered  to  compression  is  not  the  same  as  that  offered 
to  elongation,  then  the  neutral  plane  will  not  fall, 
midway  between  the  two  surfaces. 

Let  L0  be  the  original  length  of  the  beam,   and  R 
the  radius  of  curvature  of  the  neutral  plane,  then 

L0=R0; (13) 

where  6  is  the  angle  at  the  centre. 

The   length   of   the   outermost   fibre,    after   bending, 

becomes 

L  =  (R  +  d)  6 

=  R6+dd (14) 

Subtracting  equation  (13)  from  equation  (14),  we 
obtain  the  elongation;  i.e., 

e  =L-L0  =  dO (15) 

But,  by  definition 

f-'-^-:.  (16) 


ELASTICITY  107 

where  s  is  the  stress  per  unit  area.  Substituting  in 
equation  (16),  the  values  of  L0  and  e,  as  given  by 
equations  (13)  and  (15),  we  obtain 

_  sRO  _  s_R 
**     '    dd         ~d> 
from  which 


which   is    the    expression  for  the   stress  per   unit   area 
on  the  concave  and  convex  surfaces. 

In   like    manner    the    stress    per    unit    area,    at    any 
distance  x  from  the  neutral  plane,  is 


That  is,  the  stresses  at  any  section  vary  directly  as 
the  distances  from  the  neutral  plane;  and,  since  they 
have  opposite  signs  on  opposite  sides  of  the  neutral 
plane,  it  follows  that  they  constitute  a  torque,  or 
turning  moment,  about  the  neutral  axis.  The  neutral 
axis  is  denned  by  the  intersection  of  the  neutral  plane 
with  the  section  under  consideration. 

Let  the  beam  whose  width   is  b  and  depth   2  d  be 
divided,   as  in  Fig.   36,   into  a  number  of  sections  of 

width    b   and    indefinitely    small    depth    —  ,     such    that 

the  stress  throughout  the  depth  of  the  section  may  be 
considered  constant.  If  s  is  the  stress  per  unit  area 


io8 


ELEMENTARY  MECHANICS 


on    the    outermost    fibre,    or    "skin    stress,"    then    the 
stress  for  unit  area  at  a  distance  x  from  the  neutral 


FIG.  36. 

axis    is,    s  — ;    and    the    stress    for    the   element   whose 
a 

width   is   b,   and   depth   -,    at   a   distance   x   from   the 

n 

neutral  axis,  is 


where  a  is   the  area  of   the  section;    and   the  turning 

moment  of  the  section  is 

sx2 

m  =  —  —  a. 
a 

If,  now,  we  denote  the  distances  from  the  neutral 
axis  to  the  various  elements  by  xlt  x2,  x3,  etc.,  the 
areas  for  the  corresponding  sections  by  an  a2,  a3,  etc., 
and  by  M  the  turning  moment  of  the  whole  section, 
the  turning  moment,  on  one  side  of  the  neutral  axis, 
then  becomes 


M 

—  =  m 


2  -f 


x\  a2  + 


ELASTICITY  109 


that  is, 


—  =  -(x\  a  i  +  x\  a2  +  x\  «3  +  ......  +  x2n  an)  ; 

2        a 

and  the  turning  moment  for  the  whole  section  is, 
M  =    —.  (x\  al  +  x\  a2  +  x\  a3*+  ......  +  x*n  an) 


(19) 


2  I  x2  a  denotes  the  result  obtained  by  multiplying 
each  elementary  area  by  the  square  of  its  distance 
from  the  axis.  It  is  the  importance  of  the  area  with 
respect  to  the  neutral  axis,  and  may  be  appropriately 
called  moment  of  area.  In  most  text-books  it  is  called 
moment  of  inertia  and  designated  by  /.  This,  how^ 
ever,  is  not  a  well-chosen  expression,  since  it  has 
nothing  to  do  with  inertia.  To  distinguish  moment 
of  area  from  moment  of  inertia  we  shall  denote  the 
former  quantity  by  I  A.  We  then  have 

M  =  I  IA  ......     (20) 

Substituting  in  equation  (20),  the  value  of  -  as 
obtained  from  equation  (17),  we  have 


The   moment    of   area    may   readily   be    found    for   all 
regular  figures  by  integration. 


110  ELEMENTARY   MECHANICS 

It  is  possible  to  find  this  quantity,  in  certain  cases, 
without  the  aid  of  the  calculus;  but  all  such  methods 
are  cumbersome.  As  a  matter  of  illustration,  the 
moment  of  area  of  a  rectangular  figure  of  width  B 
and  depth  2  d,  will  here  be  determined  about  an  axis 
lying  in  the  plane  of  the  figure  and  half  way  between 
the  top  and  bottom  edges. 

In  Fig.  37,  let  O  O  be  the  axis,  and  assume  the 
rectangle  to  be  divided  into  an  indefinitely  large 
number  of  strips  of  equal  width,  whose  edges  are  all 
parallel  to  OO;  then  the  width 
of  each  strip  is  indefinitely  small 

d 

and    equal    to    — ,    where    n    is 

o  » 

the    number    of    strips    for    the 

half  rectangle. 

The   moment   of  area  of  any 

strip  at   a   distance   x  from   the 

axis  is,  by  what  has  been  previously  shown,  -  equal  to 
the  product  of  area  and  square  of  distance  from 
axis;  in  symbols 

ia  =  B  -  x2', (22) 

n 

and  the  moment  of  area,  of  half,  the  rectangle,  be- 
comes 

where  xlt  x2,  x3,  etc.,  are  the  respective  distances  for 
the  ist,  2d,  3d,  etc.,  strips  from  the  axis. 


ELASTICITY  1  1  1 

Substituting  for  xlt  x2,  x3,  etc.,  their  values;  namely: 

d    2  d   3  d 

-,  —  ,  —  ,  etc.,  equation  (23)  becomes 

n     n      n 

IA  =  EjL  (<!L    i^2     2_12  rc2</2  \ 

""  "  I         o    ""  iT""  ..         T~    •    *    •    •       i 

2          n     \n2        n2          n2  n2    J 

—p(i  +  4  +  9  +  .-..  +  O-      •      •      (24) 
Now,    the    series,     i2  +  22  +  32  +   ......  +  n2,     is 

equal  to    -  -    (n  +  i)  (  n  +  --  J.       But,  if  n  is  indefi- 

nitely  large,    all    quantities    such   as    i    and    ;-  vanish 

with  respect  to   it,   and  the  sum  of  the   series  equals 

n3 

—  .       Substituting    this    value,    in    equation    (24),    we 

o 

obtain 


and  the  moment  of  area  of  the  whole  rectangle,  with 
respect  to  O  O  as  an  axis,  is 

2Bd* 

J-=    ......  (25) 


Deflection  of  a  Rectangular  Bar  Clamped  at  one 
End.  Assume  a  bar  of  length  L,  width  B,  and  depth 
D,  to  be  clamped  rigidly  at  one  end  and  have  applied 
to  it  a  force  F,  at  the  free  end  normal  to  the  surface, 
as  depicted  in  Fig.  38. 

From  equation  (21),  it  is  seen  that  the  radius  of 
curvature  of  a  bent  beam  varies  inversely  as  the  turn- 


112 


ELEMENTARY   MECHANICS 


ing  moment.  But  at  any  section  of  a  beam  that  is 
in  equilibrium,  the  turning  moment  due  to  the  inter- 
nal forces  must  be  equal  to  the  bending  moment  due 
to  the  external  forces.  Now,  the  bending  moment  for 
a  beam  fastened  at  one  end,  and  an  applied  force  at 


FIG.  38. 

the  other,  varies  directly  as  the  distance  from  the 
free  end.  It  therefore  follows  that  the  curvature 
varies;  and  is  zero  at  the  free  end  and  a  maximum 
at  the  clamped  end.  For  a  distance  x  from  the 
applied  force,  we  have  for  the  radius  of  curvature 


Fx 


(26) 


ELASTICITY  113 

Assume  the  beam  to  be  divided  into  a  number  of 
equal  lengths,  each  equal  to  -  ;  n  being  an  indefi- 

nitely large  number,  so  that  the  curvature  may  be 
assumed  constant  for  each  element. 

The  angle  at  the   centre  for  an  element  at  a  dis- 
tance x  from  the  applied  force  is 

o  ..  JL     FxL. 

nRx      UIJ.IA 

If  the  total  deflection  is  small,  then  the  deflection 
due  to  an  element  is  very  small  and  equal  to  the 
product  of  the  angle  and  distance  from  the  free  end;  i.e., 

8*  =  e*x  =       .....  (27) 


But   the   total   deflection   is   equal   to   the   sum  of   the 
partial  deflections;    that  is 

d  =  d,  +  d2  +  d3  +  ......  +  dn; 

where  d^  d2,  d3,  etc.,  are  the  deflections  due  to  the 
elements  at  distances  xlt  x2)  x3,  etc.,  from  the  free 
end.  Hence,  we  have 


d  =  --  (oc\  +  x\  +  x\  +  .......  +  oc\).    .    (28) 

n  fj.  IA 

Now     #„     x2,     x3,     etc.,     are    respectively    equal    to 

L    iL  $L 

—  ,  —  ,  -  —  ,  etc.;    making   this   substitution,    in    equa- 
n     n      n 

tion  (28),  we  have 

FL        V       4*      9*     , 

5  ~r        ~  ~r        r~  ~r  ...... 


n  //  IA  V  n2        n-          n 

8 


114  ELEMENTARY   MECHANICS 

from  which 
FL3 


(29) 


But,  as  previously  shown,  when  n  is  indefinitely  large 
the  sum  of  the  series,  in  equation  (29),  becomes  equal 

n3 
to  —  .      Making  this  substitution,  we  finally  obtain 


Equation    (25)    gives   /^  =         —  .      In    deducing   this 

v) 

formula  the  depth  of  the  beam  was  designated  by  2  d; 
if,  in  place  of  this,  the  depth  be  designated  by  D, 
equation  (25)  becomes 

BD* 


Substituting,    in    equation    (30),    for   IA    its    value,    as 
given  in  equation  (31),  we  have 


(32) 


This  shows  that  when  the  deflections  are  small,  they 
vary  directly  as  the  force  and  the  cube  of  the  length, 
and  inversely  as  the  width  and  the  cube  of  the  depth. 
These  deductions  are  fully  verified  by  experiment. 

Consider,  now,  a  rectangular  bar  of  length  L,  width 
B,  and  depth  D,  supported  at  both  ends  and  a  force 
F  applied  at  its  middle  section  producing  a  deflection 


ELASTICITY 

3.  Since  the  bending  moments  at  any  two  sections 
of  the  bar,  on  opposite  sides  of  the  middle  section, 
and  at  equal  distances  from  the  point  of  support  are 
equal,  it  follows  that  the  curve  assumed  by  the  bar 
is  symmetrical  with  respect  to  the  middle  section. 

This  is  obvious  from  Fig.  39,  L  being  the  length  of 
the  bar,  and  the  force  F  being  applied  at  a  distance 

-  from    either    point  of   support,   it    follows    that    the 


F 

U-—  -4.  -> 

L_ 

2 

2 

"^~~~^—                         r)\ 

1 

FIG.  39. 


reaction   on   either   support   is   — ;    and   therefore,   the 

bending  moments,  at  equal  distances  from  the  points 
of  support,  are  equal.  Furthermore,  the  tangent  to 
the  curve  at  the  middle  section  is  parallel  to  the 
original  position  of  the  bar.  The  deflection,  therefore, 

is  the  same  as  would  be  produced  if  the  bar  were 

•p 
clamped   in  the  middle  and  subjected   to   a   force    - 

at  the  end;   the  length  of  the  bar  being  — .      Making 


Il6  ELEMENTARY     MECHANICS 

these   substitutions   in   equation    (32),   i.e.,   substituting 

77  T 

for  F,   —  ,  and  for  L,   —,  we  obtain 
2  2 

F  U 
=  ^BD* 

By  equation  (31) 

12  I  A    =  BD\ 

Substituting,  in  equation   (33)  ,  for  B  D3  its  value,  we 

have 

F  Is 
8  =  -V  .....     (34) 


Equations  (30)  and  (34)  show  that,  other  things  being 
equal,  the  deflection  varies  inversely  as  the  moment 
of  area  of  the  section.  But,  since  the  moment  of 
area  of  a  section  is  found  by  taking  the  sum  of  the 
products  obtained  by  multiplying  each  elementary  area 
by  the  square  of  its  distance  from  the  neutral  axis,  it 
follows  that  for  any  given  sectional  area,  the  moment 
of  area  may  be  increased,  and  the  deflection  decreased, 
by  distributing  the  material  in  such  a  manner  that 
the  greater  part  of  it  is  at  a  maximum  distance  from 
the  neutral  axis.  It  is  for  this  reason  that  a  plate 
will  support  a  greater  load  turned  edgewise  than  when 
lying  flat;  and  for  the  same  reason  an  I  beam  of 
given  sectional  area  will  support  a  greater  load  than 
a  rectangular  beam  of  equal  sectional  area. 

Elasticity    of   Volume.      Liquids    differ   from    solids, 
since  solids  offer  resistance  to  changes  in  form;    and 


ELASTICITY  117 

liquids,  such  as  water,  gasoline,  alcohol,  etc.,  offer 
practically  no  resistance  to  changes  in  form;  but  they 
do,  like  solids,  offer  resistance  to  changes  in  volume. 
As  an  example,  water  always  takes  the  form  of  the 
containing  vessel;  but,  to  bring  about  a  diminution 
in  volume  without  change  of  temperature,  a  pressure 
must  be  applied.  Experiment  shows  that  the  change 
per  unit  volume  is  directly  proportional  to  the  applied 
pressure;  i.e.t 


where  V  is  the  original  volume,  v  the  diminution  in 
volume,  and  p  the  applied  pressure.  Rewriting,  we 

have 

p       pV 

-  =  u  a  constant; 

v_        v 

V 

where  p  is  defined  as  the  modulus  of  voluminal 
elasticity.  For  water,  the  modulus  of  voluminal 
elasticity  is  found  to  be  about  300,000  Ibs.  per  sq.  in.; 
whereas,  for  steel,  the  modulus  of  tractional  elasticity 
is  about  28,000,000  Ibs.  per  sq.  in. 

Gases.  Like  liquids,  gases  offer  no  resistance  to 
changes  in  form;  but  differ  from  liquids,  inasmuch  as 
a  liquid  merely  takes  the  form  of  the  containing 
vessel  and  a  gas  tends  to  fill  the  whole  space  in 
which  it  is  enclosed.  That  is,  as  the  pressure  on  a  gas 
is  decreased,  the  volume  continually  increases,  and 


Il8  ELEMENTARY     MECHANICS 

finally,  if  the  pressure  be  made  indefinitely  small,  the 
volume  becomes  indefinitely  large.  At  constant  tem^ 
perature,  for  the  so-called  permanent  gases,  such  as 
air,  hydrogen,  oxygen,  etc.,  the  volume  varies  inversely 
with  the  pressure;  or,  in  other  words,  the  product 
of  pressure  and  volume  equals  a  constant.  This  is 
known  as  Boyle's  law.  In  symbols 

pv  =  k; (35) 

where  p  is  the  pressure,  v  the  corresponding  volume, 
and  k  a  constant,  whose  numerical  value  depends 
upon  the  units  chosen. 

Assume  now,  the  temperature  remaining  constant, 
that  the  pressure  receives  an  indefinitely  small  incre- 
ment A  p,  in  consequence  of  which  the  volume  suffers 
a  change  equal  to  -  A  v;  hence,  since  the  product 
of  pressure  and  volume  is  constant,  we  have 

(p  +  A  p)  (v  —  A  7;)  =  k. 
Expanding 

pv  —  p  A  v  +  v  &p  —  A  p  A  v  =  k.  .  (36) 
Subtracting  equation  (35)  from  equation  (36),  and 
rearranging,  we  obtain 

V  ^~  =  p  +    A  p (37) 

A  V 

If  we  had  assumed  a  decrement  in  pressure  and  a 
consequent  increment  in  volume,  we  would  have  found 
the  following: 

*-#-  A  P (38) 


ELASTICITY  II9 

Now,  the  nearer  A  p  and  consequently  A  v  approach 
zero  for  their  values  the  nearer  the  left-hand  members 
of  equations  (37)  and  (38)  approach  the  ratio  of  the 
change  in  pressure  to  the  corresponding  change  per 
unit  volume;  and  in  the  limit,  just  as  the  pressure  is 
beginning  to  suffer  a  change,  the  right-hand  members 
are  equal  to  each  other,  and  necessarily  equal  to  p, 
and  the  left-hand  members  are  rigidly  equal  to  the 
ratio  of  change  in  pressure  to  the  corresponding  change 
per  unit  volume.  Therefore,  the  modulus  of  vo- 
luminal  elasticity  of  a  gas  obeying  Boyle's  law  is 
numerically  equal  to  the  pressure. 


CHAPTER  XII 


STATICS 

IT  was  stated,  in  Chap.  IV,  when  dealing  with  the 
principle  of  moments,  that  experiment  shows,  that  the 
tendency  of  a  given  force  to  produce  rotation  about 
an  axis  is  independent  of  the  point  of  application,  but 
depends  solely  upon  the  intensity  of  the  force  and  its 
arm.  Assume,  as  depicted  in  Fig.  40,  the  two  non- 
parallel  coplanar  forces  a  b,  and  c  d,  applied  to  a 
rigid  body  perfectly  free  to  move;  the  points  of  ap- 
plication being  a  and  c.  Since  the  tendency  of  a 

force  to  produce  rotation  is 
not  altered  by  shifting  the 
point  of  application  along  the 
line  of  direction  of  the  force, 
the  two  forces,  a  b  and  c  d, 
may  be  replaced,  respectively, 
by  the  forces  O  g  and  O  h] 
where  O  g  =  a  b,  and  O  h  = 
c  d,  and  O  is  the  point  of  in- 
tersection of  b  a  and  d  c 

produced.  Now,  since  neither  force  has  an  arm 
with  respect  to  an  axis  through  O,  there  can  be 
no  rotation  about  this  axis.  If  then  a  third  force, 
lying  in  the  same  plane  equal  in  magnitude  to  Of, 


FIG.  40. 


STATICS  121 

which  is  the  vector  sum  of  a  b  and  c  d,  be  ap- 
plied along  the  line  Of,  but  opposite  in  direction, 
the  tendency  of  the  forces  a  b  and  c  d  to  produce 
linear  acceleration  will  be  balanced.  Since  the  three 
forces  have  no  tendency  to  produce  either  an  angular 
acceleration  about  an  axis  through  O,  or  a  linear 
acceleration  of  the  point  O,  the  three  forces  are  in 
equilibrium.  If,  on  the  other  hand,  a  force  equal  to 
Of,  and  opposite  in  direction,  whose  line  of  direction 
does  not  pass  through  the  point  O,  be  applied  to  the 
body,  there  will  be  no  tendency  to  produce  linear 
acceleration;  but  there  will  be  a  tendency  to  produce 
angular  acceleration,  since  the  two  forces  constitute 
a  couple.  Hence,  for  three  non-parallel  coplanar 
forces  to  be  in  equilibrium,  their  lines  of  direction 
must  intersect  at  a  common  point,  and  the  intensity  of 
any  one  of  the  three  forces  must  be  equal  and  opposite 
to  the  vector  sum  of  the  two  remaining  forces. 

Force  Polygon.  It  was  shown,  in  Chapter  III, 
that  the  resultant  of  two  or  more  concurrent  coplanar 
forces  may  be  found  by  vector  addition.  That  is,  if 
we  begin  at  any  point  O,  and  draw,  assuming  a  cer- 
tain scale,  a  line  in  the  direction  of  one  of  the  forces, 
and  from  the  terminal  of  this  line  draw  a  second  line, 
representing  to  the  same  scale  and  in  a  proper  direc- 
tion a  second  force,  and  from  the  terminal  of  this 
second  line,  a  third  line  representing  in  a  similar 
manner  a  third  force,  and  continue  in  this  manner 


122  ELEMENTARY    MECHANICS 

until  all  the  forces  have  been  represented,  the  line 
then  joining  the  point  O  and  the  terminal  of  the  last 
line  drawn,  represents  in  direction  and  magnitude  the 
resultant  of  all  the  forces.  And  for  the  system  of 
forces  to  be  in  equilibrium,  a  force  equal  in  magnitude 
to  the  resultant  and  opposite  in  direction  must  be 
applied  on  the  line  of  direction  of  the  resultant. 
From  this,  it  follows  that  if  a  number  of  coplanar 
forces  are  in  equilibrium,  and  a  vector  diagram  be 
drawn,  as  just  described,  the  resulting  figure  is  a 
closed  polygon.  This  may  be  further  illustrated  as 

follows:      Assume  the  forces,  Flt  F2)  F3, F7, 

when   plotted    as   just   described,    to   form   the    closed 

polygon  O,  p,  q, u,  as  depicted  in  Fig.  41. 

The  line  Rlt  joining  the  points  O  and  q,  represents 
the  resultant  of  the  two  forces  F±  and  F2,  and  there- 
fore may  replace  these  two 
forces;  similarly,  the  line  R2 
represents  the  resultant  of 
Rl  and  F3,  and  therefore 
may  replace  them.  Con- 
tinuing in  this  manner  there 
finally  remain  the  three 
forces,  R4,  F^  and  F^  but 
the  resultant  of  R4  and  F6 

is  equal  and   opposite   to  F7  and   passes   through   the 
point  O;    hence  the  system  is  in  equilibrium. 
Funicular    Polygon.       The    figure    assumed    by    a 


STATICS 


123 


closed  flexible  cord  when  in  equilibrium  under  the 
application  of  a  number  of  coplanar  forces  is  termed 
a  funicular  polygon.  Let,  as  in  Fig.  42,  the  forces 

Flt  F2,  F3, F7,  which  are  in  equilibrium,  be 

applied  to  a  closed  flexible  cord  in  such  a  manner 
that  the  cord  assumes  the  form  of  a  polygon,  O,  p, 
q, u.  The  two  following  statements  may 


FIG.  42. 

then  be  made,     (i)  The  system  being  in  equilibrium, 

the    applied    forces,    Flt    F2,    F3, F7,    must 

give  a  closed  polygon.  (2)  Since  all  points  of  applica- 
tion are  in  equilibrium,  the  vector  sum  of  the  three 
forces  at  these  points  must  be  zero;  i.e.,  the  applied 
force  F1  must  be  equal  and  opposite  to  the  resultant 
of  the  two  stresses  in  the  cord,  viz.,  Sl  and  57;  like- 
wise, the  applied  force  F2  must  be  equal  and  opposite 


1 24  ELEMENTARY     MECHANICS 

to  the  resultant  of  the  stresses,   St  and  52;    likewise 
for  F3,  etc. 

If  now,   in  Fig.   43,   we  lay  off  the  force  Fly  then 

51  and  S7  must  form  a  triangle  i  O  7  with  Ft;    laying 
off  from  the  terminal  of  Fl  the  force  F2,  then  6^  and 

52  must   combine  with  it  to  form  the  triangle   2  O  i. 
Hence  the  two  triangles  i  O  7  and  2  O  i  have  the  side 
Sl  in  common.      Likewise,  if  we  lay  off  the  force  F3 
from  the  terminal  of  F2  and  combine  with  it  the  two 
stresses  S2  and  S3  we  obtain  the  triangle  302,  having 
the  side  S2  in  common  with  the  triangle  2  O  i.     Pro- 
ceeding in  this  manner  it  is  found  that  each  triangle 
has  one  side  in  common  with  the  triangle  preceding; 
hence,  since  the  force  polygon  closes,  the  lines  drawn 
from  the  points   i,   2,   3,   etc.,   parallel  to  the  stresses 
Slt  S2,  S3,  etc.,  must  meet  in  a  point  O,  termed  the 
pole;    and  the  lengths  of  the  lines  radiating  from  the 
pole  determine  the  stresses  in  the  sides  of  the  original 
polygon. 

A  little  consideration  will  show  that  if  the  applied 
forces  acting  upon  a  closed  cord  are  in  equilibrium 
and  their  directions  and  magnitudes  are  known,  then 
the  assumption  of  the  directions  of  two  consecutive 
sides  of  the  polygon  determines  the  directions  and 
stresses  for  the  whole  polygon;  also,  if  the  shape  of 
the  polygon  and  the  directions  of  the  applied  forces 
are  given,  then  the  assumption  of  the  magnitude  of 
one  of  the  applied  forces  determines  all  the  others. 


STATICS 

The  -foregoing  demonstration  holds  equally  well  for 
an  articulated  frame,  when  in  equilibrium  under  the 
action  of  forces  applied  to  the  joints.  The  only 
difference  being  that  some  or  all  of  the  members  may 
be  under  compression  instead  of,  as  in  the  case  of 
the  cord,  under  tension. 

From  what  has  been  said  about  the  force  polygon, 
it  is  clear,  that  if  the  forces  acting  on  a  frame  are 


FIG.  44. 

not  in  equilibrium,  then  the  closing  side  of  the  poly- 
gon determines  the  direction  and  magnitude  of  the 
equilibrant. 

Assume,   as   in  Fig.   44,   the   frame  O,  p,  q,  r,  s,  t, 
u,   in  equilibrium  under  the  action  of  the   forces  Flt 

F2,   F3, F7 ;     and    further,    that    two    of   the 

sides,  such  as  q  r  and  /  s,  are  cut  across.  Since  the 
forces  FU  F2,  F3,  F6,  and  F1  are  in  equilibrium  with 
the  two  stresses  acting  along  q  r  and  s  t,  it  follows 


1 26  ELEMENTARY    MECHANICS 

that  the  resultant  of  these  two  stresses  is  the  equili- 
brant  of  all  the  forces  applied  to  the  left  of  a  b\  viz., 
Flt  F2,  F3,  FQ,  and  F7,  and  must  lie  on  a  line  passing 
through  the  point  i,  the  intersection  of  q  r  and  /  s 
produced.  If  we  now  draw  the  triangle  of  forces  for 
the  two  stresses  53,  S4,  and  the  force  F4,  and  the 
triangle  of  forces  for  the  two  stresses  S5,  S4,  and  the 
force  F5,  the  side  S4  will  be  in  common,  and  there- 
fore the  two  stresses  S3  and  S5  are  balanced  by  the 
two  forces  F±  and  F5.  Hence,  the  resultant  of  S3 
and  55  has  the  same  magnitude  and  line  of  direction 
as  the  resultant  of  F4  and  F&  and  its  line  of  direction 
must  pass  through  the  point  j,  the  intersection  of  the 
lines  of  direction  of  F4  and  F5. 

If  we  have  a  given  system  of  forces,  such  as  Flt 
F2j  F3,  FQ,  and  F7,  which  are  not  in  equilibrium,  the 
magnitude  and  line  of  direction  of  the  equilibrant  can 
readily  be  determined  as  follows:  Consider  the  given 
forces  applied  to  the  joints  of  an  articulated  frame 
and  assume  the  directions  of  two  consecutive  sides  of 
the  frame,  such  as  O  p  and  O  u,  this  determines,  as 
previously  stated,  the  directions  and  stresses  for  all 
the  members  of  the  frame;  hence,  S3  and  55  are 
known.  But,  as  has  just  been  shown,  the  resultant 
of  S3  and  55  is  the  equilibrant  of  the  given  forces. 
If  it  develops  that  S3  and  S5  are  nearly  parallel,  so 
that  it  is  impracticable  to  get  the  point  of  intersection, 
then  if  we  assume  q  i  and  /  *  to  be  cut  by  a  third 


STATICS  127 

member,  such  as  r  sy  under  an  assumed  stress  -5*4,  the 
two  forces  to  be  applied  at  the  joints  r  and  s,  namely 
F4  and  F5,  may  be  found.  For,  since  S3  and  S4  are 
given  in  both  magnitude  and  direction,  F4  is  deter- 
mined; and  similarly  F5.  But,  as  has  been  shown  in 
the  previous  demonstration,  the  resultant  of  F4  and 
FS  is  the  equilibrant  of  the  given  forces.  The  re- 
sultant of  the  given  forces  may,  of  course,  be  found, 
both  in  magnitude  and  direction,  by  the  polygon  of 
forces;  but  this  does  not  give  the  line  of  direction. 

Parallel  Forces.  If  the  forces  applied  to  the  joints 
of  an  articulated  frame  are  parallel,  then  the  force 
polygon  reduces  to  a  straight  line,  and  necessarily, 
to  be  in  equilibrium,  the  algebraic  sum  of  the  forces 
must  be  zero. 

Assume,  as  in  Fig.  45,  the  three  parallel  forces 
FU  F2,  and  .F3  applied  to  the  jointed  frame  at  the 
points  b,  cy  and  d\  and  further,  that  the  frame  is 
supported  at  the  points  a  and  e  by  the  reactions 
Rl  and  R%  parallel  to  the  applied  forces.  The  con- 
ditions here  represented  are  similar  to  a  chain  or 
cable  supporting  weights.  O  p  q  is  the  ray  polygon 
obtained  by  constructing  the  triangle  of  forces 
for  the  points  d,  c,  and  6,  as  previously  described. 
Since  the  point  e  is  in  equilibrium  the  three  forces, 
S5)  54,  and  R2  must  combine  to  form  a  triangle;  hence 
by  drawing  O  s,  in  Fig.  46,  parallel  to  a  e,  in  Fig.  45, 
the  stress  S&  is  determined  by  the  length  of  the  line 


128 


ELEMENTARY     MECHANICS 


O  s,  and  the  reaction  R2  by  the  length  of  the  line  s  p. 
Similarly  the  reaction  Rl  is  determined  by  the  length 
of  the  line  q  s.  Again,  since  one  of  the  three  forces, 
acting  at  any  point,  is  vertical,  the  horizontal  com- 
ponents of  the  stresses  for  the  two  adjacent  members 
must  be  equal  and  opposite.  But,  since  all  the  ap- 
plied forces  are  vertical,  it  follows  that  the  horizontal 


11 


FIG.  46. 


component  for  the  stresses  throughout  the  frame  "is 
constant,  and  is  determined  by  the  length  of  the  line 
t0)  in  Fig.  46;  i.e.,  the  normal  from  O  to  the  line  p  q, 
called  the  "polar  distance." 

Uniform  Horizontal  Loading.  If  a  perfectly  flexible 
cord,  supported  at  two  points,  has  applied  equal 
weights  uniformly  distributed  between  the  points  of 
support,  and  the  weight  of  the  cord  is  negligible  in 
comparison  with  the  weights,  the  curve  assumed  by 
the  cord  is  a  parabola.  The  curve  which  a  perfectly 


STATICS 


129 


flexible  non-stretchable  cord,  supported  at  two  points, 
assumes  under  its  own  weight  is  a  catenary.  The 
equation  of  this  curve,  however,  cannot  be  deduced 
without  the  aid  of  the  calculus.  If,  however,  the 
deflection  is  small  in  comparison  with  the  length  be- 
tween supports,  as  is  the  case  in  a  belt  or  cable  drive, 
such  that  we  may,  without  appreciable  error,  assume 
uniform  horizontal  loading,  the  equation  is  readily 
determined. 

Assume,  as  in  Fig.  47,  the  half  span  x,  to  be 
divided  into  n  equal  parts,  and  the  deflection  y,  of 
the  cord  a,  b,  c,  to  be  so  small,  in  comparison  with 


FIG.  47. 

the  span,  that  the  weight  of  the  cord  for  equal  hor- 
izontal distances  is  practically  constant  throughout.  If 
x  be  divided  into  an  indefinitely  large  number  of 

*V 

parts,  such  that  --  is  very  small,  then  in  any  triangle, 

such  as  d  ef,  the  chord  and  tangent  practically  coin- 
cide, and  the  deflection  is  given  by 


/y» 


y  —  —  tan  Ol  •]  —  tan  02  + 


/V* 

tan  6 
n 


-  (tan  6l  +  tan  62  +  ......  +  tan  0n).   .   (i) 


130  ELEMENTARY     MECHANICS 

Since  all  the  applied  forces,  namely  the  weights  of 
the  various  elements  of  the  cord,  act  vertically,  the 
horizontal  tension  must  be  constant  throughout.  The 
vertical  tension  at  any  point  is  equal  to  the  weight 
of  the  cord  included  between  the  point  b  and  the 
point  under  consideration.  If  the  tension  in  the  cord, 
at  any  point,  be  resolved  into  its  two  components,  the 
horizontal  component  will  be  a  constant,  which  we 
will  denote  by  t0,  and  the  vertical  will  be  as  just 
stated,  the  weight  of  the  cord  from  the  point  b  to  the 
point  under  consideration.  But  by  assumption,  we 
have  constant  loads  for  equal  horizontal  distances; 
hence,  for  the  point  /,  the  vertical  component  is 

n  ' 

where  w  is  the  weight  per  unit  length.  The  horizontal 
component  being  constant  and  equal  to  t0,  we  have 

for  the  slope  at  /,  ~x 

tan  03  =  w^— . 
nt0 

In  a  similar  manner,  we  find 

tan  6*  =  w  — ,  tan  62  =  w  — ,  tan  6n  =  w  — . 
n  t0  nt0  nt0 

Substituting    the    values    of   the    tangents    in    equation 

iu  x 

(i),  and  factoring  the  common  part  — -,  we  obtain 

n  t0 

w  oc2 

WX*     (»  +  !)«..  (2) 


2W2  t. 


STATICS  131 

X 

It  is,  of  course,  obvious  that  the  smaller  the  length  - 

becomes,  and  consequently  the  larger  n  becomes,  the 
nearer  equation  (2)  represents  the  exact  conditions. 
Assume  n  so  large  that  i  vanishes  in  comparison  with 
it;  equation  (2)  then  becomes 

w  x2 


which  is  the  equation  of  a*  parabola. 

We   will   now   determine  t0  in     terms    of    the    total 
span    5,    the    corresponding    deflection    D,    and    the 


weight    per    unit    length.       Since,    in    a    parabola,   the 
subtangent  is  bisected  at  the  vertex,   we  have  in  the 

*  Equation  (3)  is  easily  found  by  integration.  For  any  point  whose 
co-ordinates  are  x  and  y,  the  vertical  component  is  w  x;  and  the  hori- 
zontal component  being  to,  we  have 

dy      w  x 

tan  Ox  =  ~- +  =  — ; 
dx        t0 

from  which 

wx2  , 

y  =  2~T  +  c- 

2  1 0 

The  constant  of  integration  being  found  to  be   zero  from  the  condition 
that  x  =  o,  when  y  =  o. 


132  ELEMENTARY     MECHANICS 

triangle  O  g  c,   in  Fig.   48,  gb   equal  to   bO;    i.e.,  Og 
equal  to   2  D.     Hence,   we  have 


tan  8  = 


But,    since    the    slope    is   also    equal    to    the    vertical 
component   divided   by   the   horizontal   component,    we 

have 

4  D       w  S 
tan  6  =  -1—  =  - 

S          2t0y 

from  which 


Three  Forces  Meeting  in  a  Point.  Problems  whose 
solutions  involve  the  principle  that  three  coplanar 
forces  to  be  in  equilibrium  must  have  their  lines  of 
direction  meet  in  a  point,  being  of  such  frequent 
occurrence  it  will  be  well  to  consider  a  few  concrete 
cases.  The  simplest  case  is  that  of  a  weight  sup- 
ported as  shown  in  Fig.  49.  The  three  forces  meeting 
in  the  point  c  being:  the  tension  in  the  member 
supporting  the  weight  W,  and  the  stresses  in  the 
members  a  c  and  b  c,  a  c  being  under  compression  and 
b  c  under  tension.  The  point  c  being  in  equilibrium, 
the  vector  sum  of  the  two  stresses  along  a  c  and  b  c 
must  be  equal  and  opposite  to  W,  as  shown  by  the 
triangle  of  forces  c  d  e  ;  where  c  d  and  d  e  are  respect- 
ively the  reactions  of  the  members  b  c  and  a  c.  Since 
the  triangles  a  b  c  and  e  c  d  are  similar,  it  follows  that 


STATICS 


133 


the  stresses  in  the  members,  a  c,  a  b,  and  b  c,  are  to 
each  other  directly  as  the  sides  of  the  triangle  formed 
by  the  members.  But  the  stress  in  a  b  is  necessarily 


FIG.  49. 

equal   to    W.      Hence,    denoting   the   stress   in   a  c   by 
Slt  and  in  b  c  by  S2,  we  have 

W  :  a  b  :  :  Sl  :  a  c, 


from  which 


Similarly 


n (5) 


ab' 


•      (6) 


The  same  results  will,  of  course,  be  obtained  if  the 
problem  be  solved  by  the  principle  of  moments. 
Taking  moments,  about  the  point  a,  we  find 

S2Xag  =  WXac (7) 


134  ELEMENTARY     MECHANICS 

But,  the  triangles  a  b  c  and  g  a  c  are  similar;    hence 

a  c  X  a  b 

ag  =   -bT-' 

Substituting  this  value  of  a  g  in  equation  (7),  we  find 


which  is  the  same  as  equation  (6).      Similarly,  taking 
moments  about  the  point  b,  we  find 

Sl  X  ab  =  W  X  ac, 


from  which 


-  W  — 
l  ab' 


which  is  the  same  as  equation   (5). 

The  moment  tending  to  turn  the  support  b  m,  in  a 
clockwise  direction,  about  the  point  m,  is 

W  X  a  c  =  S2  X  m  i  -  5X  h. 

In  order  to  balance  this  moment  about  the  point  m, 
a  tie  rod  b  k  may  be  used  which  must  be  under  a 
tension  53,  such  that 

S3Xmj  =  WXac  =  S2Xmi-S1  h. 

Bar  Supported  by  a  Horizontal  and  Vertical  Surface. 

Assume  a  bar  resting  with  one  end  on  a  horizontal 
surface,  and  the  other  against  a  vertical  surface,  in 
such  a  manner  that  it  lies  in  a  plane  normal  to  the 
two  surfaces.  We  have  here  three  forces;  i.e.,  the 
reactions  of  the  two  surfaces,  and  a  force,  equal  to 
the  weight  of  the  bar,  applied  at  its  centre  of  gravity 


STATICS 


135 


acting  vertically  downward.  Remembering,  that  when 
there  is  no  friction,  the  reactions  must  be  normal  to 
the  surfaces,  it  follows  that  equilibrium  cannot  obtain 
for  perfectly  smooth  surfaces;  for,  in  such  a  case, 
the  lines  of  direction  of  twp  of  the  forces  are  parallel 
and  the  third  acts  at  right  angles  to  them.  If,  how- 
ever, the  horizontal  surface  is  rough,  equilibrium  will 
obtain,  providing  the  normal  reaction  of  the  vertical 
surface  is  not  greater  than  the  force  of  friction  on 
the  horizontal  surface.  Assume,  as  depicted  in  Fig. 
50,  the  bar  a  b,  whose  weight  is  W,  and  whose  centre 
of  gravity  is  at  G,  having  the  end  a  resting  against 
a  perfectly  smooth  vertical  surface,  and  the  end  b, 
resting  upon  the  rough  horizontal  surface  O  b.  The 
vertical  surface,  O  a,  being  perfectly  smooth,  the  re- 


FIG.   50. 

action  Rt  must  be  normal  to  it.  Hence,  for  equili- 
brium to  obtain,  the  line  of  direction  of  R2  must  pass 
through  the  point  c,  the  intersection  of  Rl}  and  W. 


136  ELEMENTARY    MECHANICS 

The  magnitude  of  W  being  known,  the  two  reactions 
are  found  by  constructing  the  parallelogram  b  d  e  f. 
It  is  evident  from  the  figure  that  as  the  angle  ft  de- 
creases, R!  increases;  but,  for  the  equilibrium  to  obtain, 
the  balancing  force  due  to  friction  must  be  equal  and 
opposite  to  Rv.  Since  the  greatest  value  the  force  of 
friction  can  have'  is  the  product  of  weight  and  co- 
efficient of  friction,  it  follows  that  when  the  angle  ft 
has  been  decreased  to  a  value  such  that 


sliding  will  be  impending.  For  all  values  of  ft  less 
than  this,  equilibrium  is  impossible.  To  determine 
the  critical  value  for  the  angle  /?,  i.e.,  that  value 
when  sliding  is  impending,  take  moments  about  the 
point  of  support  b.  For  equilibrium  to  obtain,  we 

must  have 

Wmcosft=Rll  sin  /?;...      (9) 

where  m  is  the  distance  from  the  centre  of  gravity  to 
the  point  of  support  6,  and  /  the  length  of  the  bar. 
From  equation  (9),  we  find 

Wm 
tam-'jtf 

and   substituting  for  Rlt   its   value,   as  obtained   from 

equation  (8),  we  obtain 

m  .    . 

tan  ft  =  —  ......     (10) 

f*-  1 

For  all  values  of  ft  greater  than  that  given  by  equa- 
tion (10),  equilibrium  will  obtain. 


PROBLEMS 


CHAPTER  I 

1.  A  body  moving  uniformly  passes  over  a  distance  of  10 
feet  in  2  seconds.     What  is  its  speed  ?     How  long  .will  it  take 
to  travel  25  feet? 

Ans.  5  ft.  per  sec.  ;  5  sec. 

2.  A  particle  has  a  uniform  speed  of  30  kilometers  per  day. 
How  long  will  it  take  to  travel  3,500  millimeters? 

Ans.  10.08  sec. 

3.  A  body  starts  from  rest  with  a  constant  acceleration  of 
10  ft.  per  sec.  per  sec.     Determine  the  distance  passed  over  in 
the  3d,  5th,  and  yth  seconds,  and  the  total  distance  passed  over 
in  10  seconds. 

Ans.  25  ft.;  45  ft.;  65  ft.;  and  500  ft. 

4.  The  velocity  of  a  body  changes  uniformly  from  10  ft.  per 
sec.  to  25  ft.  per  sec.  in  3  seconds.     What  is  its  constant  ac- 
celeration?    When  is  its  velocity  75  ft.  per  sec.?     How  long 
will  it  have  been  in  motion,  assuming  it  to  have  started  from 
rest  ?     What  space  will  it  have  passed  over  ? 

Ans.  5  ft.  per  sec.  per  sec.;   in  10  sec.;    15  sec.;   562.5  ft. 

5.  A  body  changes  speed  from  100  meters  per  second  to  60 


138  PROBLEMS 

meters  per  second  in  going  40  meters.  What  is  the  acceleration, 
assuming  it  to  be  constant  ?  With  the  same  acceleration,  how 
far  will  the  body  have  moved  before  coming  to  rest  ?  In  what 
time  will  it  come  to  rest  ? 

Ans.  —  80  meters  per  sec.  per  sec.;   22.5  meters;  0.75  sec. 

6.  At  a  given  instant  a  body  is  found  to  have  a  velocity  of  200 
ft.  per  sec.     Ten  seconds  later  it  is  found  to  have  a  velocity  of 
500  ft.  per  sec.     What  is  its  acceleration,  assuming  it  constant  ? 
What  space  did  it  cover  in  the  10  seconds? 

Ans.  30  ft.  per  sec.  per  sec.;  3,500  ft. 

7.  A  body  starting  from  rest  with  a  uniformly  accelerated 
motion  passes  over  a  distance  of  36  kilometers  in  2  hours. 
What  is  its   acceleration   in   cm.    per  sec.   per  sec.?      What 
was  its   velocity,   and  how  far  had  it  travelled,  15  minutes 
after  starting  ? 

Ans.  5/36  cm.  per  sec.  per  sec.;    125  cm.  per  sec.;    562.5 
meters. 

8.  The  velocity  of  a  particle  changes  uniformly  from  30  ft. 
per  sec.  to  20  ft.  per  sec.  in  passing  over  25  ft.     What  is  its 
acceleration  ?     How  long  will  it  be  before  coming  to  rest,  and 
what  distance  will  it  have  traversed  in  that  time,  if  its  retarda- 
tion is  constant? 

Ans.  —  10  ft.  per  sec.  per  sec.;  2  sec.;   20  ft. 

9.  With  what  acceleration,  and  how  far  must  a  body  move  to 
have  a  speed  of  30  m.  p.  h.  in  30  seconds  after  starting  from 
rest?      What   retardation    would   destroy    this   speed    in    10 
seconds  ?     How  far  would  the  body  have  travelled  ? 

Ans.  1.467  ft.  per  sec.  per  sec.;  1/8  mile;  4.4  ft.  per  sec.  per 
sec.;    1/24  mile. 


PROBLEMS  139 

10.  A   body  moving    with  a  speed  of    40    m.  p.  h.  is  re- 
tarded  uniformly  and  brought  to  rest    in   500  ft.     What  was 
the  retardation  in  miles  per  hour  per  sec.,  and  in  feet  per  sec. 
per  sec.  ? 

Ans.  2.346;  3.44. 

11.  What  is  the  curvature  of  a  circle  whose  diameter  is  eight 
feet? 

Ans.  1/4  radian  per  foot. 

12.  The  direction  of  motion  of  a  particle  is.changed  uniformly 
by  0.25  radians  in  passing   over  5  feet.     What  curve  has  it 
described  and  what  are  its  dimensions? 

Ans.  Circle;  40  ft.  diameter. 

13.  What  will  be  the  change  in  direction  of  a  particle  moving 
10  ft.  in  the  circumference  of  a  circle  100  ft.  in  diameter? 

Ans.  11.46°. 

14.  What  distance  has  a  body  moved  in  the  circumference  of 
a  circle  of  25  ft.  radius,  if  its  change  in  direction  of  motion  was 
0.6  radians  ? 

Ans.  15  ft. 

15.  A  rotating  disc  makes  3,000  r.p.m.     Find  its  angular 
velocity.     Find  the  linear  speed  of  a  point  2  ft.  from  the  axis  of 
rotation. 

Ans.  100  TT  radians  per  sec. ;   200  TT  ft.  per  sec. 

16.  The  linear  speed  of  a  point  on  a  rotating  body  is  90  ft. 
per  minute  and  its  distance  from  the  axis  of  rotation  is  5  ft. 
How  long  will  it  take  the  body  to  sweep  out  54  radians  ? 

Ans.  3  minutes. 


140  PROBLEMS 

17.  A  bucket  is  raised  by  a  rope  passsing  over  a  sheave  near 
the  end  of  a  derrick  boom,  at  the  uniform  speed  of  ten  feet  per 
second.     If  the  rope  winds  up  on  a  drum  four  feet  in  diameter, 
then,  neglecting  the  thickness  of  the  rope,  what  is  the  angular 
velocity  of  the  drum?     If  the  given  velocity  were  acquired  in 
three  seconds,  what  would  be  the  angular  acceleration  of  the 
drum? 

Ans.  5  rad.  per  sec.;   1.67  rad.  per  sec.  per  sec. 

18.  A  flywheel  starting  from  rest  is  found  in  twenty  seconds 
to  be  revolving  150  times  per  minute.     What  is  its  angular 
acceleration,   assuming  it   a  constant?     What   would   be  its 
angular  velocity  at  the  end  of  one  minute  from  rest  ? 

Ans.  7T/4  rad.  per  sec.  per  sec.;  15  it  rad.  per  sec. 

19.  A  rotating  body,  having  an  angular  acceleration  of  10  rad. 
per  sec.  per  sec.,  has  been  in  motion  10  sec.     What  is  its  angular 
velocity  and  how   many  rotations  has  it  made?    What  time 
will  elapse  and  how  many  rotations  will  be  made,  before  its 
angular  velocity  is  900  rad.  per  sec.? 

Ans.  100  rad.  per  sec. ;  250/71-;  80  sec.;  2o,ooo/;r. 

20.  A  rotating  body  starting  from  rest  has  been  in  motion 
n  seconds.     If  its  angular  acceleration  is  a  how  many  rotations 
will  be  made  during  the  next  /  seconds  ? 

Ans.  - '- —  (2n  +  /). 
47: 


CHAPTER  II 

1.  Add  together  the  four  vectors,  ten  north,  fifteen  east, 
seven  south,  and  twelve  west. 

Ans.  3\/2  N.E. 

2.  Resolve  the  vector  twenty  into  two  vectors  making  angles 
of  30°  and  60°  on  each  side  of  it. 

Ans.  io\/3;  10. 

3.  Resolve  the  vector  A  into  the  vectors  B,  C,  D,  E,  and  F; 
having  assumed  the  directions  of  the  vectors  B,  C,  D,  and  E. 

4.  Two  cars  are  moving  along  level  lines  inclined  at  an  angle 
of  45°,  with  speeds  of  20  and  30  m.p.h.     If  the  cars  are  re- 
spectively 500  and  600  feet  from  the  crossing  point,  show  by 
diagram  the  motion  of  each  car  as  it  appears  to  the  driver  of 
the  other. 

5.  A  captain  wished  to  sail  a  ship,  whose  speed  is  12  m.p.n. 
in  a  Southeast  direction.     There  is  a  current  running  5  m.p.h. 
which  sets  the  ship  due  West,  off  her  course.     In  what  direction 
must  she  be  headed  in  order  to  sail  in  the   S.E.   direction? 
Show  by  diagram. 

6.  In  problem  (5)  what  will  be  the  actual  speed  of  the  ship 
in  the  S.E.  direction  ?     If  the  ship  had  been  headed  S.E.,  where 
would  she  have  been  at  the  end  of  one  hour  ? 

Ans.  7.93  m.p.h.;    9.17  miles  from  starting  point,  and  3.54 
miles  out  of  her  course. 

I4.I 


I42  PROBLEMS 

7.  A  boat  steams  across  a  river  at  right  angles  to  the  course  of 
the  river  with  a  speed  of  10  m.p.h.     If  the  boat  reaches  the 
opposite  shore  2  miles  below  the  starting  point,  and  if  the  river 
is  4  miles  wide,  what  was  the  speed  of  the  current? 

Ans.  5  m.p.h. 

8.  A  point  moves  in  the  circumference  of  a  circle,  whose 
diameter  is  20  ft.,  with  a  uniform  angular  velocity  of  5  radian 
per  sec.     What  is  the  centripetal  acceleration  ? 

Ans.  250  ft.  per  sec.  per  sec. 

9.  If  in  problem  (8),  the  point  started  from  rest  and  moved 
with  a  uniform  angular  acceleration  of  2  radians  per  sec.  per 
sec.,  what  would  be  the  centripetal  acceleration  at  the  end  of  five 
seconds  ? 

Ans.  1,000  ft.  per  sec.  per  sec. 

10.  If  a  point  moves  with  a  uniform  speed  of  10  ft.  per  second 
in  the  circumference  of  a  circle  of  diameter  40  feet,  what  will  be 
the  velocity  and  the  acceleration  of  the  projection  of  the  point 
upon  a  diameter,  when  the  point  has  moved  through  71/4  radians 
from  the  extremity  of  the  diameter? 

Ans.  5\/2  ft.  per  sec.;  5/^2  ft.  per  sec.  per  sec. 


CHAPTER  III 

i.  A  picture  whose  weight  is  21  Ibs.  is  suspended  by  a  cord 
hung  over  a  peg.  Each  branch  of  the  cord  makes  an  angle  of 
30°  with  the  vertical.  What  is  the  tension  in  the  cord? 

Ans. 


2.  An  inclined  plane  has  a  rise  of  i  in  10.     Assuming  no 
friction,  what  force,  acting  parallel  to  the  plane,  will  just  support 
a  weight  of  100  Ibs.  ?     What  force  parallel  to  the  base  ? 

Ans.  9.95  Ibs.;  10  Ibs. 

3.  A  body  weighing  50  Ibs.  rests  on  a  plane  inclined  at  an 
angle  of  30°.     Assuming  the  coefficient  of  friction  to  be  0.3, 
what  force,  acting  parallel  to  the  plane,  will  draw  the  body  up 
the  plane  with  uniform  motion  ? 

Ans.  37.99  Ibs. 

4.  A  body  weighing  100  Ibs.  rests  upon  a  plane  inclined  at 
an  angle  of  45°.     Assuming  the  coefficient  of  friction  to  be  o.i, 
what  force,  parallel  to  the  base,  will  draw  the  body  up  the  plane 
with  uniform  motion? 

Ans.  122.2  Ibs. 

5.  A  mass  of  i  gram,  perfectly  free  to  move,  starts  from  rest 
under  the  action  of  a  constant  force  of  i  dyne.     In  what  time 
is  i  erg  of  work  performed  ? 

Ans.  ^/2  sec. 

6.  A  mass  suspended  from  a  railway  car  by  a  cord  3  ft.  long, 
rises  a  vertical  height  of  o.i  inches  upon  starting.    If  the  mass 


144  PROBLEMS 

is  in  equilibrium  in  this  position,  what  is  the  acceleration  of  the 
car? 

Ans.  28.68  inches  per  sec.  per  sec. 

7.  A  mass  of  10  pounds,  resting  on  a  horizontal  plane,  is 
moved  8  feet  in  8  seconds,  starting  from  rest.     What  force, 
parallel  to  the  plane,  was  necessary  if  the  coefficient  of  friction 

is  0.3  ? 

Ans.  3.08  Ibs. 

8.  A  weight  of  100  Ibs.  rests  on  a  plane,  inclined  sin~ 1  0.6 
with  the  horizontal.     What  happens,  respectively,  when  forces 
of  20,  40,  80,  and  100  Ibs.  are  applied  to  the  body  up  and 
parallel  to  the  plane?     Coefficient  of  friction  0.25. 

9.  If  a  carriage  be  slipped  from  a  train  moving  at  30  m.p.h. 
up  a  plane  inclined  sin—1  0.02  with   the  horizontal,  how  far, 
friction  being  neglected,  will  it  move  before  beginning  to  run 
back? 

Ans.  1512.5  ft. 

10.  A  mass  of  10  pounds  is  whirled  in  a  horizontal  plane  by 
a  cord  10  feet  long  capable  of  carrying  but  50  Ibs.     How  many 
revolutions  per  minute  are  necessary  to  break  it  ? 

Ans.  38.2. 

1 1 .  400  masses  of  6  pounds  each  are  distributed  around  the 
circumference  of  a  rotating  body  at  a  mean  distance  of  3  ft. 
from  the  axis.     What  will  be  the  tension  in  a  cord  wrapped 
round  them  when  the  system  is  making  200  r.p.m.  ? 

Ans.  31,416  Ibs. 

12.  A  4oo-ton  train  travels  round  a  curve  i  mile  in  radius  at 


PROBLEMS  145 

40  m.p.h.     What  is  the  horizontal  component  of  the  pressure 
on  the  rails? 
Ans.  16,300  Ibs. 

13.  If  the  centre  of  gravity  of  the  train  in  the  preceding  prob- 
lem be    midway  between   the   rails  (5    ft.    gauge),  and  5    ft. 
above  them,  what  must  be  the  speed  so  that  the  train  is  on  the 
point  of  turning  over? 

Ans.  198.17  m.p.h. 

14.  Assuming,  in  problem  (12),  the  centre  of  gravity  midway 
between  rails,  then  how  much  would  it  be  necessary  to  incline 
the  track  in  order  that  there  be  equal  pressure  on  them  ? 

Ans.  tan"1©. 02037. 

15.  Find  the  number  of  vibrations  that  would  be  executed 
per  minute  by  a  mass  of  5  pounds  attached  to  a  spring,  obeying 
Hooke's  Law,  if  a  weight  of  4  Ibs.  causes  an  elongation  of  10 
inches. 

Ans.  53  (very  nearly). 

1 6.  A  mass  of  5  pounds  when  attached  to  a  spring  obeying 
Hooke's  Law  executes  240  vibrations  in  3  minutes.     What  is 
the  force  required  to  elongate  the  spring  i  ft.  ? 

Ans.  ii  Ibs.  (very  nearly). 

17.  A  mass  of  5  pounds  attached  to  a  spring  obeying  Hooke's 
Law,  vibrates  with  an  amplitude  of  2.5  ft.,  executing  100  vibra- 
tions in  3  minutes.     When  the  displacement  is  18  inches,  find 
the  value  of  the  acceleration,  velocity,  kinetic  energy,  potential 
energy,  and  total  energy. 

Ans.  18.3  ft.  per  sec.  per  sec.;    6.98  ft.  per  sec.;    121.9  ft. 
poundals;  68.5  ft.  poundals;  190.4  ft.  poundals. 


146  PROBLEMS 

1 8.  Find  the  value  of  the  quantities  named  in  the  preceding 
problem  when  the  mass  is  at  a  position  such  that  0.15  seconds 
elapse  before  reaching  the  equilibrium  position. 

Am.  15.23  ft.  per  sec.  per  sec.;    7.56  ft.  per  sec.;    142.8  ft. 
poundals;  47.6  ft.  poundals. 

19.  The  velocity  of  a  body  moving  with  a  S.H.M.  is  20  ft. 
per  sec.  when  3  ft.  from  the  equilibrium  position,  and  15  ft. 
per  sec.  when  4  ft.  from  it.     What  are  the  maximum  values  of 
the  displacement,  velocity,  and  acceleration? 

Am.  5  ft.;  25  ft.  per  sec.;   125  ft.  per  sec.  per  sec. 

20.  A  weight  of  100  Ibs.  rests  on  a  platform  which  moves 
with  an  S.H.M.,  having  an  amplitude  of  4  ft.,  and  a  period  of 
4  seconds.     When  the  platform  is  2  ft.  above  the  equilibrium 
position  and  moving  upward,  what  is  the    pressure    exerted 
by  the  weight?     What  is  the  pressure,  at  the  same  position, 
when  moving  downward? 

2 1 .  How  long  will  it  take  a  force  of  i  ,000  Ibs.  to  stop  a  200- 
ton  mass  moving  at  60  m.p.h.?     What  work  will  be  done?      / 

Ans.  18  min.  20  sec.;   24.44  H.P.  hours. 

22.  What  is  the  constant  force  required  to  stop  a  mass  of 
200  tons,  moving  at  60  m.p.h.,  in  loofeet?     How  much  work 
has  been  done  ? 

Am.  242  tons;  484  X  io5  ft.  Ibs. 

23.  A  weight  of  500  Ibs.  rests  upon  a  plane  having  an  in- 
clination of  30°.     If  the  coefficient  of  friction  is  o.i,  how  much 
work  will  be  done  in  drawing  the  weight,  with  uniform  speed, 
io  ft.  up  the  plane? 

Ans.  2,933  ft- 


PROBLEMS  147 

24.  A  mass  of  10  grams,  perfectly  free  to  move,  is  acted  upon 
for  5  seconds  by  a  constant  force  of  50  dynes.     What  kinetic 
energy  will  the  mass  have  at  the  end  of  10  seconds? 

Ans.  3,125  ergs. 

25.  A  weight  of  200  Ibs.  falls  from  a  height  of  15  feet  upon 
the  head  of  a  pile,  which  under  the  action  of  the  blow,  sinks  3 
inches  into  the  ground.     What  was  the  resistance  ? 

Ans.  12,000  Ibs. 


CHAPTER  IV 

1.  A  force  of  5  poundals  acts  upon  a  mass  of  10  pounds  for 
2  minutes.     How  much  will  the  momentum  of  the  body  be 
changed  ? 

Ans.  600  F.P.S.  units. 

2.  What  force  in  5  seconds  will  change  the  speed  of  a  100 
gram  mass  from  40  cm.  per  sec.  to  100  cm.  per  sec.  ? 

Ans.  1,200  dynes. 

3.  Masses  of  5  and  10  pounds,  having  velocities  of  8  and  5 
ft.  per  second  respectively,  collide.     What  are  their  velocities 
after  impact  if  the  coefficient  of  restitution  is  unity?     What 
if  0.5?     What  if  zero?     Illustrate  conditions  by  diagrams. 

4.  Solve  problem  (3)  when  the  lo-pound  mass  has  a  velocity 
of  -  5  ft.  per  sec. 

5.  Solve  problem  (4)  when  the  5 -pound  mass  has  a  velocity 
of  10  ft.  per  sec. 

6.  Solve  problem  (5)  when  the  lo-pound  mass  has  a  velocity 
of  —  5  ft.  per  sec. 

7.  Solve  problem  (3)  when  the  5 -pound  mass  has  a  velocity 
of  12  ft.  per  sec. 

8.  Solve  problem  (7)  when  the  lo-pound  mass  has  a  velocity 
of  —  5  ft.  per  sec. 

148 


PROBLEMS  149 

9.  An  inelastic  mass  impinges  directly  upon  another  twenty 
times  as  great.     What  was  its  initial  velocity  if,  after  impact, 
both  move  a  distance  of  3  feet  in  2  seconds  ? 

Ans.  31.5  ft.  per  sec. 

10.  A   one-ounce  bullet    is    fired  with    a  velocity  of  1,600 
ft.  per   sec.    from   a    2o-pound   rifle,    which    is   held   against 
a  mass   of    180   pounds.     With   what   velocity   did  the   rifle 
"  kick  "back? 

Ans.  6  inches  per  sec. 

11.  A  body  is  dropped  from  a  height  of  16  ft.  and  bounces  a 
height  of  9  ft.     What  is  the  coefficient  of  restitution?     To 
what  height  will  the  body  bounce  the  next  time  ? 

Ans.  0.75;  5TV  ft. 

12.  A  jet  of  water  from  an  orifice  i  sq.  inch  in  section  im- 
pinges against  a  wall.     What  is  the  force  exerted  if  1 20  gallons 
are  delivered  per  minute? 

Ans.  20.07  Mb8- 

13.  What  force  is  exerted  upon  a  gun  delivering  200  one- 
ounce  bullets  per  minute  with  a  speed  of  1,600  ft.  per  sec.? 

Ans.  10.42  Ibs. 

14.  A  projectile  whose  mass  is  100  pounds  is  fired  into  a 
target,  whose  mass  is  20,000  pounds,  with  a  velocity  of  1,000 
ft.  per  sec.     If  the  target  be  free  to  move,  find  the  loss  in  energy 
during  impact. 

Ans.  1,555,000  ft.  Ibs. 

15.  A  square  board  2  feet  on  a  side  and  weighing  3  Ibs.,  has 
placed  at  the  corners  A,  B,  C,  and  D,  weights  of  i  lb.,  2  Ibs. 


150  PROBLEMS 

3  Ibs.,  and  4  Ibs.,  respectively.     Where  must  the  board  be 
supported  in  order  to  remain  in  a  horizontal  position  ? 

Ans.  1.307  ft.  from  the  side  A  B,  i.o  ft.  from  the  side  A  D. 

16.  A  uniform  bar  10  ft.  long  and  weighing  10  Ibs.,  has 
weights  of  5,  6,  7,  and  8  Ibs.  suspended  from  it  at  distances  of 
i,  2,  3,  and  4  feet  respectively,  from  one  end.  Find  the  equili- 
brant. 

Ans.  36  Ibs.  acting  upward  3  ft.  4  inches  from  the  end  from 
which  measurements  were  taken. 


CHAPTER  V 

MOMENTS  OF  INERTIA 

I.  Hollow  cylinder  of  mass  Af,  length  /,  internal  radius  r^ 
and  external  radius  r2.   Moment  of  inertia  with  respect  to  axis 

M 

of  cylinder  =  —  (r2l  +  r22). 

II.  Solid  sphere,  mass  M,  and  radius  r.     Moment  of  inertia 
with  respect  to  a  diameter  =  —Mr2. 

III.  Cone,  mass  M,  radius  r,  and  height  h  in  the  direction 

of  axis.    Moment  of  inertia  with  respect  to  its  axis  =  -  -  . 

10 

IV.  Rectangular  plate,  mass  M,  length  /,  width  6,  and  depth 
d,  in  the  direction  of  axis.     Moment  of  inertia  with  respect 
to  a    perpendicular    axis    passing    through  centre  of   mass 


V.  Thin  rectangular  plate,  mass  M,  length  /,  and  breadth 
b,  in  direction  of  axis.     Moment  of  inertia  with  respect  to  one 

MP 

end  as  an  axis  =  -  . 
3 

VI.  Thin  triangular  plate,  mass  M,  altitude  h,  and  base  b. 

Moment  of  inertia  'with  base  as  axis  =  -  7—  . 

6 

VII.  Moment  of  inertia  of  a  thin  plate  with  respect  to  an 
axis  normal  to  the  figure  is  equal  to  the  sum  of  the  moments  of 
inertia  with  respect  to  two  axes,  coplanar  with  the  figure,  at 
right  angles  to  each  other,  and  whose  intersection  is  coincident 
with  the  normal  axis. 

VIII.  To  show  that  the  acceleration  of  a  body  rolling,  on  a 


152 


PROBLEMS 


circular  section,  down  an  inclined  plane,  is  less  than  g  sin  6; 
where  6  is  the  angle  of  inclination  of  the  plane.  Let,  as  de- 
picted in  the  accompanying  figure,  the  homogeneous  body  roll 
on  the  circular  section  whose  centre 
is  at  C;  and  let  its  moment  of  inertia, 
about  an  axis  normal  to  this  section, 
and  passing  through  the  centre  of 
mass  C,  be  MK2',  where  M  is  the 
mass  of  the  body  and  K  the  radius 
of  gyration  for  this  particular  axis. 
The  moment  of  inertia  then,  about  the  parallel  instantaneous 
axis,  through  the  point  of  contact  p,  is 

7  =  M  (K2  +  r2). 

The  torque,  about  the  axis  through  p,  is 
G  =  Mgr  sin  6; 

and  since  angular  acceleration  is  given  by  the  ratio  of  torque 
to  moment  of  inertia,  we  have 

M  g  r  sin  0  r 


Multiplying  by  r,  we  find  for  the  linear  acceleration 

which  shows,  since       2       2    is  less  than  unity,  that  a  is  less 

than  g  sin  6.  It  furthermore  shows,  since  M  is  eliminated, 
that  the  acceleration  is  independent  of  the  mass;  and  since,  in 
any  case  K  =  kr;  where  k  is  some  constant,  we  may  write 

r*  I 

—*g  sin  0  = 


a  = 


gsm 


k2  -fi 

which  shows  that  the  acceleration  is  independent  of  the  radius 
of  the  section  on  which  the  body  rolls. 

It  will  prove  instructive  to  the  student  to  demonstrate  these 
results  from  the  principle  of  energy. 


PROBLEMS  153 

1.  Find  the  moment  of  inertia  of  a  thin  square  plate  with 
respect  to  a  diagonal  as  axis. 

2.  Find  the  moment  of  inertia  of  a  thin  trapezoidal  plate 
about  its  base  as  axis. 

3.  Find  the  moment  of  inertia  of  a  thin  circular  plate  with 
respect  to  a  diameter  as  axis. 

4.  Find  the  moment  of  inertia  of  a  thin  iron  plate  (density 
=  480  pounds  per  cu.  ft.),  3  feet  long  and  i  square  inch  in  cross- 
section,  about  one  end. 

Ans.  30  pound  ft.2 

5.  Find  the  moment  of  inertia  of  a  plate  whose  mass  is  10 
pounds,  the  dimensions  being  five  feet  by  two  feet  by  one-eighth 
inch,  about  an  axis  through  the  centre  of  mass  parallel  to  the 
five-foot  side. 

Ans.  31/3  pound  ft.2 

6.  What  is  the  moment  of  inertia  of  a  thin  plate  of  iron,  10 
feet  long  and  i  square  inch  in  cross-section,  about  an  axis  in  the 
plane  of  the  plate,  parallel  to  a  short  edge,  and  2  feet  from  it, 
assuming  the  density  of  iron  to  be  480  pounds  per  cubic  foot? 

Ans.  577.8  pound  ft.2 

7.  What  is  the  moment  of  inertia  of  a  thin  rectangular  plate, 
2  feet  by  6  feet,  whose  mass  is  4  pounds,  about  a  long  edge? 
About  a  short  edge?     About  an  axis  through  the  centre,  per- 
pendicular to  its  plane  ? 

Ans.  5  1/3  pound  ft.2;  48  pound  ft.2;   13  1/3  pound  ft.2 

8.  A  wheel,  consisting  of  a  solid  disk  of  stone,  4  feet  in 


154  PROBLEMS 

diameter  and  6  inches  thick,  makes  120  revolutions  per  minute. 
If  the  density  of  the  stone  is  150  pounds  per  cubic  foot,  what 
kinetic  energy  does  the  wheel  possess? 

Am.  2,400  TT?  ft.  Ibs. 

9.  A  cast-iron  flywheel  has  a  rim  i  inch  thick,  12  inches  wide, 
and  4  ft.  mean  diameter;  6  spokes,  19.5  inches  long  and  4  inches 
by  3  inches  in  section.  The  hub  is  10  inches  external  diameter, 
4  inches  internal  diameter,  and  10  inches  thick.  What  is  the 
moment  of  inertia  of  the  flywheel,  the  density  of  cast  iron  being 
480  pounds  per  cubic  foot  ? 

Mass.  I. 

Rim 502  pounds  2,008  pound  ft.2 

6  spokes.  .  .  .  390  pounds  679  pound  ft.2 

Hub 183  pounds  37  pound  ft.2 


Total 1,075  pounds  2,724  pound  ft.2 

10.  A  hollo wr  cylinder  6  inches  long,  is  free  to  vibrate  about  a 
knife-edge  support,  passing  through  it.     If  its  external  diameter 
is  36  inches  and  its  internal  diameter  is  18  inches,  what  is  the 
moment  of  inertia  about  its  support  ?     Density  of  material  400 
pounds  per  cubic  ft. 

Ans.  Mass,  1,060  pounds;  /,  2,087  pound  ft 2 

11.  A  body  free  to  rotate  about  an  axis  has  its  speed  changed 
from  900  r.p.m.  to  600  r.p.m.  in  90  rotations.     If  its  moment 
of  inertia  is  4,000  pound  ft.2,  what  constant  torque  brought 
about  the  change? 

Ans.  545.4  Ib.  ft. 

12.  What   is  the   moment   of    inertia  of   a   body   free    to 
rotate,    if    a   constant   torque   of    4,000   Ib.    ft.    is    necessary 


PROBLEMS  155 

to   produce  a  speed  of  1,200  r.p.m.,  in  90  seconds,  starting 
from  rest? 

Ans.  91,670  pound  ft.2 

13.  What  is  the  constant  torque  required  to  stop  a  body, 
whose  moment  of  inertia  is  500  pound  ft.2,  making  1,200 r.p.m., 
in  60  rotations? 

Ans.  327  Ib.  ft. 

14.  A  rotating  body  mounted  on  a  shaft,  4  inches  in  diameter, 
is  making  240  r.p.m.  and  is  being  retarded  by  the  friction  of  the 
bearings  which  support  it.     The  coefficient  of  friction  is  o.oi, 
mass  of  rotating  system   2,500  pounds,   and  its  moment  of 
inertia  7,500  pound  ft.2     What  time  elapses  before  coming  to 
rest  ?     How  much  work  has  been  done  ? 

Ans.  23  min.  33.7  sec.;  74,000  ft.  Ibs. 

15.  A  flywheel,  \vhose  mass  is  2,000  pounds  and  radius  of 
gyration  3  ft.,  takes  2  minutes  to  come  to  rest  from  a  speed  of 
240  r.p.m.     What  is  the  retardation,  and  coefficient  of  friction 
at  the  bearings  ?     Diameter  of  shaft  4  inches. 

Ans.  0.21  rad.  per  sec.  per  sec.;  0.354. 

1 6.  What  energy  is  possessed  by  the  fly  wheel  in  problem 
(9),  if  it  makes  300  r.p.m.  ? 

Ans.  42,000  ft.  Ibs. 

17.  A  car  weighing  42  tons  including  its  8  wheels  of  500 
pounds  each,  is  moving  at  30  m.p.h.     If  the  diameter  of  each 
wheel  is  3  ft.,  and  its  radius  of  gyration  is  i  ft.,  what  is  the 
kinetic  energy  possessed  by  the  car  ? 

Ans.  2,595,000  ft.  Ibs. 


CHAPTER  VI 

1.  Assuming  an  efficiency  of  75  per  cent,  what  quantity  of 
water,  per  minute,  will  a  40  H.P.  engine  raise  from  a  mine  300 
ft.  deep? 

Am.  52.9  cu.  ft. 

2.  A  mass  of  25  kilograms,  perfectly  free  to  move,  is  under 
the  action  of  a  constant  force.     Its  velocity  changes  from  2 
meters  per  second  to  4  meters  per  second  in  passing  over  3 
meters.     Find  the  power  in  watts,  H.P.,  and  kilogram-meters 
per  sec.,  that  is  being  expended  when  the  velocity  is  4  meters 
per  sec. 

Ans.  200;  0.268;  20.4. 

3.  A  mass  of  200  tons,  having  a  velocity  of  50  m.p.h.  is  re- 
tarded uniformly  at  2   miles  per  hour  per  sec.     What  is  the 
mean  rate  in  kilowatts  at  which  its  kinetic  energy  is  destroyed  ? 
What  work  in  kilowatt-hours  will  be  performed  ? 

Ans.  1823.5;  12.66. 

4.  A  mass  of  500  kilograms,  starting  from  rest,  is  made  to 
move  with  a  uniformly  accelerated  motion  up  a  plane,  inclined 
30°  with  the  horizontal,  and  passes  over  a  distance  of  8  meters 
in  8  seconds.     If  the  coefficient  of  friction  is  0.25,  find  the  total 
work  done  during  the  eight  seconds.     What  power  is  being 
expended  at  the  end  of  the  eighth  second  ? 

Ans.  29,080  joules;  9.748  H.P. 

156 


PROBLEMS  IS7 

5.  What    power    must    be    expended    to    propel,    at    15 
m.p.h.,    a    2oo-ton    mass,   up   a    plane    inclined    with     the 
horizontal    sin"1 0.05?      What    if    the   friction    be   15    Ibs. 
per  ton  ? 

Am.  8,000  H.P.;  8,120  H.P. 

6.  To  propel  a  mass  of  400  tons  along  a  horizontal  surface  at 
60  miles  an  hour  requires  950  H.P.     What  is  the  coefficient  of 
friction  ? 

Ans.  0.0074. 

7.  The    angular  velocity  of    a  rotating    mass  changes    in 
5   seconds  from   100  radians  per  second  to  40  radians  per 
second.       If  the  mass  is   1,000    pounds  and  its    radius    of 
gyration   5   ft.,   find    the    time    rate    at    which   its    angular 
momentum  is  changing. 

Ans.  300,000  F.P.S.  units. 

8.  What  is  the  time  rate  at  which  work  is  being  done  at  the 
end  of  the  5  seconds  in  the  previous  problem  ? 

Ans.  508.6  K.W. 

9.  An  engine  is  doing  work,  at  the  rate  of  40  H.P.,  in 
maintaining    a    constant    speed    of    300  r.p.m.    against    the 
force  of  friction  applied  to  the  circumference  of  its  flywheel 
by  means  of  a  Prony  brake.     If  the  centre  of  the  flywheel 
and  the  platform  of  the  balance,  upon  which  the  lever  arm 
of   the  brake   rests,  are  in  the  same  horizontal  plane,  then 
what  will  be  the  reading  of  the  balance,  if  the  distance  between 
the  centre  of  the  flywheel  and  the  point  of  contact   on  the 
platform  is  4.5  ft.? 

Ans.  155.6  Ibs. 


158  PROBLEMS 

10.  A  force  of  50  Ibs.  friction  exists  at  the  circumference  of  a 
pulley  8  inches  in  diameter.     If  a  constant  speed  of  1,525  r.p.m. 
is  maintained,  what  is  the  H.P.  expended? 

Ans.  4.84. 

11.  If  the  pulley  in  the  previous  problem  is  hollow  and 
capable  of  containing  4  pounds  of  water,  how  long  will  it  take 
to  raise  the  water  160°  F.,  assuming  no  heat  losses? 

Ans.  3  min.  7  sec. 


CHAPTER  VII 

1.  Posts  are  placed  at  the  corners  of  a  square.     A  rope  is 
passed  completely  around  them.     In  what   direction   would 
the  posts  fall  if  unable  to  withstand  the  pressure  ? 

2.  An  elevator  car  weighing  2,000  Ibs.  is  made  to  ascend  with 
a  constant  acceleration  of  16  ft.  per  sec.  per  sec.     What  is  the 
tension  in  the  rope  hauling  the  cage?     If  the  elevator  car  were 
falling  with  a  constant  acceleration  of  32  ft.  per  sec.  per  sec., 
what  would  be  the  tension  in  the  rope  ? 

Ans.  3,000  Ibs.;  zero  Ibs. 

3.  A  100  Ib.  weight  rests  upon  the  floor  of  an  elevator  car, 
which  is  descending  with  a  constant  acceleration  of  2  ft.  per 
sec.  per  sec.     What  pressure  does  the  weight  exert  upon  the 
floor  ?     If  the  elevator  car  were  ascending  with  a  uniform  speed 
of  1 6  ft.  per  sec.  what  pressure  would  the  weight  exert  upon  the 
floor? 

Ans.  93.75  Ibs.;  100  Ibs. 

4.  In  an  Atwood's  Machine  a  rope  is  led  over  a  pulley  and 
has  attached  to   the  ends  masses  of  8  pounds  and  7  pounds 
respectively.     Assuming  the  equivalent  mass  of  the  pulley  to 
be  i  pound  and  neglecting  the  mass  of  the  rope,  what  is  the 
acceleration  of  the  system?    What  is  the  tension  in  each  branch 
of  the  rope? 

Ans.  2.0  ft.  per  sec.  per  sec.;  7.5  Ibs.  and  7.437  Ibs. 

5.  Masses  of  40  and  50  grams  are  connected  together  by  a 
thin  cord  and  hung  over  a  frictionless  pulley,  whose  mass  may 


160  PROBLEMS 

be  neglected.     What  distance  will  be  traversed  in  2  seconds 
when  starting  from  rest?     What  is  the  tension  in  the  cord? 

Ans.  218  cm.;   44.4  gms. 

6.  If  the  masses,  in  the  previous  problem,  had  but  passed 
over  a  distance  of  196  cm.  in  the  2  seconds,  and  the  diameter 
of  the  pulley  is  10  cm.,  what  would  be  the  moment  of  inertia 
of   the  pulley?     Its   equivalent   mass?     The   tension   in    the 
cords  ? 

Ans.  250  gram  cm.2;   10  grams;  45  gms.;  44  gms. 

7.  A  rotating  body,  together  with  the  shaft  upon  which  it  is 
mounted,  has  a  moment  of  inertia  of  5,000  pound  ft.2     What 
weight  must  be  suspended  from  a  rope  wrapped  round  the  shaft 
to  produce  a  speed  of  90  r.p.m.  in  one  minute?     Diameter  of 
shaft  12  inches. 

Ans.  49.2  Ibs. 

8.  Two  masses  of  0.4  and  0.6  pounds,  respectively,  are  sup- 
ported by  a  cord,  passing  over  a  frictionless  pulley,  whose 
radius  is  3   inches.     It  is  found  that  the  masses  in  starting 
from  rest  pass  over  a  distance  of  16  ft.  in  4  seconds.     What 
is  the  moment  of  inertia  of  the  pulley?     What  is  its  equiv- 
alent mass  ? 

Ans.  0.1375  pound  ft.2;  2.2  pounds. 

9.  The  coefficient  of  friction  between  a  mass  of  10  pounds 
and  a  horizontal  plane  is  0.2.     The  mass  starting  from  rest 
moves  over  a  distance  of  24.6  ft.  in  2  seconds,  and  is  propelled 
by  a  cord  parallel  to  the  plane,  passing  over  a  pulley  6  inches  in 
diameter,  supporting  a  weight  of  12  Ibs.     What  is  the  moment 
of  inertia  of  the  pulley  ? 

Ans.  0.251  pound  ft.2 


PROBLEMS  l6l 

10.  A  mass  is  drawn  up  a  plane,  inclined  6°  with  the  hori- 
zontal, by  a  cord  parallel  to  the  plane,  passing  over  a  frictionless 
pulley  (radius  r,  moment  of  inertia  /),  suspending  a  weight  W. 
If  the  coefficient  of  friction  between  the  mass  and  plane  is  /*, 
what  is  the  acceleration?  What  are  the  tensions  in  the  cord? 

n.  A  solid  drum,  whose  mass  is  500  pounds,  has  a  diameter 
of  4  feet.  There  is  wound  about  the  drum  a  rope  supporting  a 
load  of  1,000  Ibs.  Assuming  no  friction,  what  H.P.  is  expended 
at  the  instant  the  weight  is  48  ft.  from  its  initial  position,  if  the 
time  consumed  was  4  seconds  and  the  acceleration  constant? 
What  is  the  tension  in  the  rope  ? 

Ans.  53.9;  1187.5  lbs- 


ii 


CHAPTER  IX 

1.  What  is  the  length  of  a  simple  pendulum  that  will  make 
one  oscillation  per  second  where  g  =  32  //.  per  sec.  per  sec.? 

Ans.  38.9  inches. 

2.  What  will  be  the  time  of  vibration  of  a  simple  penduium, 
whose  length  is  one  meter,  where  g  =  980  cm.  per  sec.  per  sec.? 

Ans.  2.007  seconds. 

3.  Find  the  time  of  vibration  of  a  thin  rod  4  ft.  long  when 
vibrating  about  an  axis  6  inches  from  one  end.     What  is  the 
equivalent  length  of  the  simple  pendulum? 

Ans.  1.717  sec.;  2.389  ft. 

4.  What  is  the  radius  of  gyration  of  the  rod  in  problem  (3), 
as  suspended? 

Ans.  1.893  ft. 

5.  What  will  be  the  minimum  time  of  vibration  of  the  bar  of 
problem  (3)  ? 

Ans.  1.688  seconds. 

6.  If  the  mass  of  the  pendulum  in  problem  (3)  is  2  pounds, 
what  are  the   masses  which,  when  concentrated  at  the  axis  of 
suspension  and  oscillation  respectively,  wrill  constitute  a  pendu- 
lum having  the  same  characteristics? 

Ans.  0.745  pounds;  1.255  pounds. 

162 


PROBLEMS  163 

7.  What  is  the  time  of  vibration  of  the  hollow  cylinder,  of 
problem  (10),  Chapter  V  ?   What  is  the  length  of  the  equivalent 
simple  pendulum? 

Ans.  1.8  seconds;   2.625  ft. 

8.  A  thin  rectangular  bar  6  ft.  long  is  suspended  by  a  cord  9 
ft.  long.     At  what  point  must  a  blow  be  struck  to  make  the 
system  vibrate  smoothly  about  the  point  of  suspension  of  the 
cord?     About  the  point  of  suspension  of  the  bar? 

Ans.  2.75  ft.  from  bottom;  2.00  ft.  from  bottom. 

9.  A  mass,  attached  to  a  cord  5  ft.  long,  rotates  in  a  horizon- 
tal plane  making  27  r.p.m.     How  high  will  the  mass  be  from  its 
position  of  rest  ?     What  will  be  the  velocity  of  the  mass  in  its 
path? 

Ans.  i  ft.;  8.48  ft.  per  sec. 

10.  What  is  the  time  of  vibration  of  a  rectangular  plate,  six 
feet  by  eight  feet,  about  one  corner,  the  axis  being  perpendicular 
to  the  plane  of  the  plate  ? 

Ans.  2.868  seconds. 

n.  Determine  where  else  the  plate  of  problem  (10)  must  be 
suspended  in  order  that  it  will  vibrate  in  the  same  time. 

Ans.  1 1  ft.  from  the  centre  of  the  plate. 

12.  A  one-pound  projectile  is  fired  into  a  suspended  block 
of  wood,  whose  mass  is  319  pounds,  and  causes  it  to  rise,  with- 
out rotation,  a  vertical  height  of  6  inches.  What  was  the 
velocity  of  the  projectile  at  the  instant  of  the  impact? 

Ans.  1,810.2  feet  per  sec. 


CHAPTER  X 

1.  A  body  falling  from  rest  passes  over  496  ft.  during  a  certain 
second.     How  long  had  it  been  in  motion? 

Ans.  15  sec. 

2.  A  freely  falling  body  starting  from  rest  has  been  in  motion 
n  seconds.     What  will  be  the  space  traversed  by  it  during  the 
next  /  seconds? 

Ans.  —  (2  n  +  t). 

3.  A  body  is  dropped  from  an  elevator  ascending  with  a 
speed  of  20  ft.  per  sec.     How  long  will  it  take  to  reach  its  highest 
point,  and  then  fall  100  ft.  ?     What  will  its  velocity  be  at  the 
end  of  that  time? 

Ans.  3J  sec.;   100  ft.  per  sec. 

4.  If  an  elevator  is  descending  with  a  speed  of  20  ft.  per  sec., 
how  long  will  it  take  a  body  dropped  from  it  to  fall  100  ft.  ? 
What  will  be  its  velocity  at  the  end  of  that  time  ? 

Ans.  1.952  sec.;  82.46  ft.  per  sec. 

5.  A  body,  projected  vertically,  has  an  upward  velocity  of 
100  ft.  per  sec.  after  being  in  motion  for  5  seconds.     How  high 
is  it  ?     How  much  further  will  it  continue  to  rise  ?     What  time 
will  elapse  before  reaching  the  ground  ? 

Ans.  900  ft.;  1 56!  ft;  nj  sec. 

164 


PROBLEMS  165 

6.  A  body  is  dropped  from  a  height  of  100  ft.  and  at  the  same 
time  another  body  is  projected  vertically  upward  with  a  veloc- 
ity sufficient    to   carry  it  to  that  point.     When  and  where  will 
the  bodies  pass  each  other  ? 

Ans.  75  ft.  from  bottom,  in  1.25  sec. 

7.  Two  masses  are  let  fall  from  the  same  place  one  second 
apart.     How  long  a  time  will  elapse  before  the  masses  are  32  ft. 
apart  ? 

Ans.  1^  seconds  after  the  first  mass  is  let  fall. 

8.  A  body  is  projected  vertically  upward  with  a  velocity  of 
40  ft.  per  sec.     To  what  height  will  it  rise  ?     How  long  will  it  be 
before  reaching  the  level  from  which  it  was  projected  ?     At  the 
instant  the  body  is  20  ft.  from  that  level  a  second  body  is  dropped 
from  there.     At  what  distance  below  the  level  will  the  bodies 
meet? 

Ans.  25  ft.;   2.5  sec.;   20  ft. 

9.  An  elevator  car  is  ascending  at  the  uniform  rate  of  32  ft. 
per  second,  and  when  240  ft.  above  the  floor  of  the  building  a 
ball  is  kicked  off.     In  what  time  will  the  ball  reach  the  floor? 

Ans.  5.0  sec. 

10.  In  problem  (9),  how  high  will  the  elevator  car  be  when 
the  ball  strikes  the  floor? 

Ans.  400  ft. 

n.  A  balloon  is  sinking  at  the  uniform  rate  of  10  ft.  per 
second.  A  ball  is  thrown  upward  with  a  velocity,  relative  to 
the  balloon,  of  74  ft.  per  second.  When  the  ball  is  at  its  highest 
point,  how  far  down  from  it  will  the  balloon  be? 

Ans.  84  ft. 


1 66  PROBLEMS 

12.  What  is  the  actual  velocity  of  a  projectile  in  its  path  at  an 
elevation  y  ? 

Ans.  (u2  -\-  v2  —  2  g  yp 

13.  A  projectile  is  shot  over  the  ocean  from  the  top  of  a  hill 
i, 600  ft.  high,  with  a  horizontal  velocity  of  1,200  ft.  per  sec. 
Neglecting  the  curvature  of  the  earth,  where  will  the  projectile 
strike  the  water  ?     How  soon  will  it  strike  the  water  ? 

Ans.  12,000  ft.  from  the  projection  of  the  cannon's  mouth 
on  the  plane  of  the  ocean;   10  sec. 

14.  A  projectile  is  shot  out  with  a  velocity  of  300  ft.  per  sec., 
and  after  travelling  1,000  ft.  arrives  at  the  same  level  with  a 
velocity  of  250  ft.  per  sec.     What  was  the  average  resistance 
of  the  air? 

Ans.  0.43  (nearly)  of  the  weight  of  the  projectile. 

15.  A  gun  is  elevated  at  an  angle  of  60°  to  the  horizon.     A 
projectile  shot  out  reaches  the  ground  in  54  1/8  seconds.     Find 
the  initial  velocity  and  range. 

Ans.  1,000  ft.  per  sec.;  27,062. 5  ft. 

1 6.  A  projectile  is  shot  out  ot  a  gun,  elevated  at. an  angle  of 
30°,  with  a  velocity  of  800  ft.  per  sec.     Find  the  time  of  flight, 
maximum  height,  and  range. 

Ans.  25  sec.;  2, 500  ft.;   17,300  ft. 

17.  With  the  same  gun  as  in  problem  (16),  but  elevated  at  an 
angle  so  as  to  give  the  maximum  range,  find  the  time  of  flight, 
height  to  which  the  projectile  will  rise,  and  the  range. 

.  35-355  sec.;  5,000  ft.;  20,000  ft. 


PROBLEMS  167 

1 8.  A  projectile  is  shot  out  at  an  angle  of  45°  to  the  horizon, 
with  a  velocity  of  1,414.2  ft.  per  sec.     To  what  height  will  it 
rise?     What  will  be  its  range ? 

Ans.  15,625  ft.;  62,500  ft. 

19.  A  baseball  is  struck   at   an  angle  which  will   give  the 
maximum  range  with  a  velocity  of  100  feet  per  second.     A 
fence  12  ft.  high  is  at  a  distance  such  that  the  ball  will  just  clear 
it.     What  is  the  distance  ? 

Ans.  300  ft.  or  12.5  ft. 

20.  A  projectile  after  being  in  motion  for  31  seconds  strikes 
the  ground  at  an  angle  of  30°.     Assuming  it  to  have  been  fired 
from  the  same  horizontal  plane,  what  was  its  velocity? 

Ans.  992  ft.  per  sec. 

21.  A  projectile  having  a  range  of  30,000  ft.  was  in  motion  50 
seconds.     If  its  initial  velocity  was  1,000  ft.  per  sec.,  how  high 
did  it  rise? 

Ans.  10,000  ft. 

22.  WThat  was  the  velocity,  in  ft.  per  sec.,  of  the  projectile  in 
the  previous  problem  at  an  elevation  of  5,000  ft.  ?     At  10,000 
ft?    After  being  in  motion  12.5  seconds?     After  25  seconds? 

Ans.  824.6;  600;   721.1;   600. 

23.  A  mass  of  25  pounds  is  whirled  in  a  vertical  plane  by  a 
string  10  ft.  long  until  it  breaks.     If  the  string  is  capable  of 
supporting  but  45  Ibs.,  and  the  centre  of  the  circle,  which  the 
mass  describes,  is  74  feet  above  a  horizontal  surface,  what  will 
be  the  range  of  the  mass? 

Ans.  32  ft. 


CHAPTER  XI 

1.  A  brass  rod  80  cm.  long  is  elongated  1.8  mm.  by  a  load  of 
400  Ibs.     If  its  diameter  is  3  mm.,  what  is  the  modulus  of 
elasticity  ? 

Ans.  1.12  X  io12  dynes  per  sq.  cm. 

2.  The  modulus  of  tractional  elasticity  of  wrought  iron  is 
15,000  tons  per  sq.  inch,  and  the  safe  working  load  for  tensile 
stresses  is  10,000  Ibs.  per  sq.  inch.     What  will  be  the  elongation 
per  foot  of  a  bar  so  loaded  ? 

Ans.  0.004  inches. 

3.  Allowing  the  elongation  per  foot  as  found  from  the  previ- 
ous problem  to  be  a  safe  working  practice,  what  must  be 
the  diameter  of  a  wrought-iron  bar  to  support  a  load  of  5  tons  ? 

Ans.  1.13  inches. 

4.  What  is  the  largest  force  that  can  be  safely  sustained  by  a 
phosphor  bronze  wire  o.  i  inches  in  diameter,  if  an  elongation  of 
0.0084  inches  per  foot  is  allowable?  ft  =  7,000  tons  per  sq.  inch. 

Ans.  77  Ibs. 

5.  Find  the  modulus  of  rigidity  of  a  steel  wire  75  cm.  long, 
4  mm.  in  diameter,  if  a  force  of  5  Ibs.,  having  a  moment  arm  of 
io  cm.,  produces  a  twist  of  47°. 

Ans,  8.09  X  io11  c.g.s.  units. 

6.  A  torque  of  i  dyne-cm,  applied  to  a  quartz  fibre,  io  cm. 
long,  produces  a  twist  of  360°.     If  the  modulus  of  rigidity  of 

1 68 


PROBLEMS  169 

quartz  is  2.9  X  iou  c.g.s.  units,  what  is  the  diameter  of  the 
fibre? 

Ans.  0.0273  mm. 

7.  Find  and  compare  the  forces  in  microdynes,  which  when 
having  a  moment  arm  of  i  cm.,  and  applied  to  quartz  fibres 
10  cm.  long,  of  diameters  0.002,  0.003,  o«oo4>  0-005?  0.006  mm., 
will  produce  twists  of  i  radian.     Modulus  of  rigidity  for  quartz 
2.9  X  io11  dynes  per  sq.  cm. 

Ans.  4.56;  23.1;  72.9;  178;  369. 

8.  Find  the  period  of  a  torsion  pendulum,  consisting  of  a 
cylindrical  mass  of  3  kilograms  io  cm.  in  diameter,  suspended 
by  a  phosphor  bronze  wire  2  meters  long,  0.5  mm.  in  diameter, 
whose  modulus  of  rigidity  is  3.6  X  io11  c.g.s.  units. 

Ans.  36.61  sec. 

9.  What  is  the  moment  of  torsion  of  the  wire  in  the  preceding 
problem  ? 

Ans.  1,104  c.g.s.  units. 

10.  Find  the  moment  of  inertia  of  the  body  which  when 
added  to  the  cylindrical  mass  of  the  torsion  pendulum,  in  the 
preceding  problem,  makes  the  period  50  seconds. 

Ans.  32,400  gram  cm.2 

11.  If  the  modulus  of  tractional  elasticity  of  the  wire,  in  the 
preceding  problem,  is  9.3  X  io11  dynes  per   sq.  cm.,  find  the 
elongation  produced  by  the  added  mass  of  3  kilograms. 

Ans.  3.22  mm. 

12.  Find  the  modulus  of  rigidity  of  the  wire  of  a  torsion 
pendulum  185  cm.  long,  0.5  mm.  in  diameter,  which  has  a 


1 70  PROBLEMS 

period  of  50  seconds;   the  rotating  mass  having  a  moment  of 
inertia  of  86,000  gram  cm.2 

Am.  4.09  X  lo11  c.g.s.  units. 

13.  A  steel  shaft  50  ft.  long,  and  3  inches  in  diameter,  trans- 
mits 10  H.P.  at  250  r.p.m.     How  much  is  it  twisted  if  its 
modulus  of  rigidity  is  6,000  tons  per  sq.  inch  ? 

Ans.  0.908°. 

14.  What  diameter  steel  shaft  is  necessary  to  transmit  50 
H.P.  at  200  r.p.m.  ?     The  permissible  twist  is  i°  for  every  20 
diameters  contained  in  the  length,  and  the  modulus  of  rigidity 
of  steel  is  6,000  tons  per  sq.  inch. 

Ans.  2.48  inches. 

15.  What  horse-power  is  being  transmitted  by  a  steel  shaft 
320  mm.   in  diameter,  making  75    r.p.m.,  if  the  amount  of 
torsion  for  a  length  of  25  meters  is  17.5  mm.,  measured  along 
the  circumference  of  the  shaft?     Modulus  of  rigidity  of  steel 
8  X  iou  c.g.s.  units. 

Ans.  3,793. 

1 6.  The  formula  for  the  angle  of  torsion  may  be  written, 

C*  T 

6  =  K  — ;r.     Find  the  numerical  value  of  K.  so  that  6  will 
nd* 

be  expressed  in  degrees  when  G  is  measured  in  Ib.ft.,  L  in  feet, 
n  in  tons  per  sq.  inch,  and  d,  the  diameter,  in  inches. 

Ans.  42.02. 

17.  The  angle  of  torsion  in  degrees  in  a  shaft,  d  inches  in 
diameter,  whose  modulus  of  rigidity  is  n  tons  per  sq.  inch,  when 


PROBLEMS 

transmitting  H.P.  horse-power,  a  distance  of  L  feet,  at  N  r.p.m., 

L  HP 
is  0°  =  C  \7    4.     What  is  the  numerical  value  of  C? 

n  N  d4 

Ans.  220,700. 

1 8.  With  the  aid  of  the  constant  found  in  problem   (17), 
recalculate  problem  (13). 

19.  A  rectangular   bar  0.75    inches   wide,  and   0.25    inches 
thick,  is  lying  flat;  when  it  is  supported  at  points  3  ft.  apart,  it 
is  deflected  0.177  inches  by  a  load  of  5  Ibs.  midway  between 
them.     What  is  the  modulus  of  elasticity? 

Ans.  14,060  tons  per  sq.  inch. 

20.  Compare  the  deflection  of  a  rectangular  beam  supported 
at  each  end  lying  flat,  with  that  obtained  when  turned  on  edge, 
other  conditions  remaining  the  same. 

21 .  A  rectangular  beam,  whose  section  is  8  inches  by  4  inches, 
is  supported  at  the  ends,  and  is  loaded  at  the  middle  by  a  force, 
applied  parallel  to  the  8-inch  side.     If  the  resulting  deflection 
be  one  inch,  what  would  the  deflection  be  if  the  same  force  were 
applied  at  the  middle,  parallel  to  the  4-inch  side  ? 

Ans.  4  inches. 

22.  A  beam  of  rectangular  cross-section,  20  feet  long,  18 
inches  deep,  and  6  inches  wide,  is  supported  at  the  ends  and 
loaded  at  the  middle  by  a  force  of  1,000  Ibs.     If  the  modulus 
of  elasticity  is  1,500,000  Ibs.  per  sq.  inch,  what  is  the  deflection? 

Ans.  0.0658  inches. 


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*  Qualitative  Chemical  Analysis 8vo,  1  25 

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Spencer's  Handbook  for  Cane  Sugar  Manufacturers 16mo,  mor.  3  00 

Handbook  for  Chemists  of  Beet-sugar  Houses 16mo,  mor.  3  00 

Stockbridge's  Rocks  and  Soils 8vo,  2  50 

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*  Tillman's  Descriptive  General  Chemistry 8vo,  3  00 

*  Elementary  Lessons  in  Heat 8vo,  1  50 

Treadwell's  Qualitative  Analysis.     (Hall.) 8vo.  3  00 

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5 


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*  Weaver's  Military  Explosives 8vo,  3  00 

Wells's  Laboratory  Guide  in  Qualitative  Chemical  Analysis 8vo,  1   50 

Short  Course  in  Inorganic  Qualitative  Chemical  Analysis  for  Engineering 

Students ,  12mo,  1  50 

Text-book  of  Chemical  Arithmetic 12mo,  1   25 

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CIVIL    ENGINEERING. 

BRIDGES    AND    ROOFS.     HYDRAULICS.     MATERIALS    OF    ENGINEER- 
ING.    RAILWAY    ENGINEERING. 

Baker's  Engineers'  Surveying  Instruments 12mo,  3  00 

Bixby's  Graphical  Computing  Table Paper  19 i  X  24  J  inches.  25 

Breed  and  Hosmer's  Principles  and  Practice  of  Surveying.      Vol.  I.  Elemen- 
tary Surveying 8vo,  3  00 

Vol.  II.      Higher  Surveying 8vo,  2  50 

*  Burr's  Ancient  and  Modern  Engineering  and  the  Isthmian  Canal 8vo,  3  50 

Comstock's  Field  Astronomy  for  Engineers 8vo,  2  50 

*  Corthell's  Allowable  Pressure  on  Deep  Foundations 12mo,  1   25 

Crandall's  Text-book  on  Geodesy  and  Least  Squares 8vo,  3  00 

Davis's  Elevation  and  Stadia  Tables 8vo,  1   00 

Elliott's  Engineering  for  Land  Drainage 12mo,  1   50 

*  Fiebeger's  Treatise  on  Civil  Engineering 8vo,  5  00 

Flemer's  Photographic  Methods  and  Instruments 8vo,  5  00 

Folwell's  Sewerage.      (Designing  and  Maintenance.) 8vo,  3  00 

Freitag's  Architectural  Engineering 8vo,  3   50 

French  and  Ives's  Stereotomy 8vo,  2  50 

Goodhue's  Municipal  Improvements 12mo,  1   50 

*  Hauch  and  Rice's  Tables  of  Quantities  for  Preliminary  Estimates..  .  12mo,  1   25 

Hayford's  Text-book  of  Geodetic  Astronomy 8vo,  3  00 

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Ives   and   Hilts's    Problems    in    Surveying,   Railroad   Surveying  and   Geod- 
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Johnson's  (L.  J.)  Statics  by  Algebraic  and  Graphic  Methods 8vo,  2  00 

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*  Mahan's  Descriptive  Geometry 8vo,  1  50 

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Rideal's  Sewage  and  the  Bacterial  Purification  of  Sewage 8vo,  4  00 

Riemer's  Shaft-sinking  under  Difficult  Conditions.  (Corning  and  Peele.).8vo,  3  00 

Siebert  and  Biggin's  Modern  Stone-cutting  and  Masonry. . . 8vo,  1  50 

Smith's  Manual  of  Topographical  Drawing.     (McMillan.) 8vo.  2  50 

6 


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Venable's  Garbage  Crematories  in  America 8vo,  2  00 

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Wait's  Engineering  and  Architectural  Jurisprudence 8vo,  6  00 

Sheep,  6  53 

Law  of  Contracts 8vo,  3  00 

Law  of   Operations   Preliminary   to   Construction   in   Engineering  and 

Architecture 8vo,  5  00 

Sheep,  5  50 

Warren's  Stereo tomy — Problems  in  Stone-cutting 8vo,  2  50 

*  Waterbury's   Vest-Pocket    Hand-book   of   Mathematics   for   Engineers. 

2|X5f  inches,  mor.  1   00 

*  Enlarppr!  Edition,  Including  Tables mor.  1   50 

Webb's  Proolems  in   the  Use  and  Adjustment  of  Engineering  Instruments. 

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BRIDGES    AND  ROOFS. 

Boiler's  Practical  Treatise  on  the  Construction  of  Iron  Highway  Bridges.. 8vo,  2  00 

*  Thames  River  Bridge Oblong  paper,  5  00 

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Du  Bois's  Mechanics  of  Engineering.      Vol.  II Small  4to,  10  00 

Foster's  Treatise  on  Wooden  Trestle  Bridges 4to,  5  00 

Fowler's  Ordinary  Foundations 8vo,  3  50 

Greene's  Arches  in  Wood,  Iron,  and  Stone 8vo,  2  53 

Bridge  Trusses 8vo,  2  53 

Roof  Trusses 8vo,  1  25 

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Heller's  Stresses  in  Structures  and  the  Accompanying  Deformations. .  .  .  8vo,  3  03 

Howe's  Design  of  Simple  Roof-trusses  in  Wood  and  Steel 8vo.  2  00 

Symmetrical  Masonry  Arches 8vo,  2  50 

Treatise  on  Arches 8vo,  4  00 

*  Jacoby's  Structural  Details,  or  Elements  of  Design  in  Heavy  Framing,  8vo,  2  25 
Johnson,  Bryan  and  Turneaure's  Theory  and   Practice  in  the  Designing  of 

Modern  Framed  Structures Small  4to,  10  00 

*  Johnson,  Bryan  and  Turneaure's  Theory  and  Practice  in  the  Designing  of 

Modern  Framed  Structures.     New  Edition.      Part  1 8vo,  3  00 

Merriman  and  Jacoby's  Text-book  on  Roofs  and  Bridges: 

Part  I.      Stresses  in  Simple  Trusses 8vov  2  50 

Part  II.    Graphic  Statics 8vo,  2  50 

Part  III.      Bridge  Design 8vo,  2  50 

Part  IV.   Higher  Structures 8vo,  2  50 

Morison's  Memphis  Bridge Oblong  4to,  10  00 

Sondericker's  Graphic  Statics,   with   Applications  to  Trusses,   Beams,   and 

Arches 8vo,  2  00 

Waddell's  De  Pontibus,  Pocket-book  for  Bridge  Engineers 16mo,  mor.  2  00 

*  Specifications  for  Steel  Bridges 12mo,  50 

Waddell  and  Harrington's  Bridge  Engineering.      (In  Preparation.) 


HYDRAULICS. 

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Bazin's  Experiments  upon  the  Contraction  of  the  Liquid  Vein  Issuing  from 

an  Orifice.     (Trautwine.) 8vo,     2  00 

Bovey 's  Treatise  on  Hydraulics 8vo,     5  00 

Church's  Diagrams  of  Mean  Velocity  of  Water  in  Open  Channels. 

Oblong  4to,  paper,      1  50 

Hydraulic  Motors 8vo,     2  00 

7 


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Flather's  Dynamometers,  and  the  Measurement  of  Power 12mo,  3  00 

Folwell's  Water-supply  Engineering 8vo,  4  00 

Frizell's  Water-power 8vo,  5  00 

Fuertes's  Water  and  Public  Health 12mo,  1   50 

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Ganguillet  and  Kutter's  General  Formula  for  the  Uniform  Flow  of  Water  in 

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Hazen's  Clean  Water  and  How  to  Get  It Large  12mo,  1   50 

Filtration  of  Public  Water-supplies 8vo,  3  00 

Hazelhurst's  Towers  and  Tanks  for  Water-works 8vo,  2  50 

Herschel's  115  Experiments  on  the  Carrying  Capacity  of  Large,  Riveted,  Metal 

Conduits 8vo,  2  00 

Hoyt  and  Grover's  River  Discharge 8vo,  2  00 

Hubbard    and    Kiersted's   Water-works    Management    and    Maintenance. 

8vo,  4  00 

*  Lyndon's    Development   and    Electrical    Distribution    of    Water    Power. 

8vo,  3  00 

Mason's  Water-supply.      (Considered    Principally   from    a    Sanitary    Stand- 
point.)   8vo,  4  00 

Merriman's  Treatise  on  Hydraulics 8vo,  5  00 

*  Molitor's  Hydraulics  of  Rivers,  Weirs  and  Sluices 8vo,  2  00 

*  Morrison  and  Brodie's  High  Masonry  Dam  Design 8vo,  1    50 

*  Richards's  Laboratory  Notes  on  Industrial  Water  Analysis 8vo,  50 

Schuyler's   Reservoirs  for  Irrigation,   Water-power,   and   Domestic   Water- 
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*  Thomas  and  Watt's  Improvement  of  Rivers 4to,  6  00 

Turneaure  and  Russell's  Public  Water-supplies 8vo,  5  00 

Wegmann's  Design  and  Construction  of  Dams.      5th  Ed.,  enlarged 4to,  6  CO 

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Whipple's  Value  of  Pure  Water Large  12mo,  1  00 

Williams  and  Hazen's  Hydraulic  Tables 8vo,  1  50 

Wilson's  Irrigation  Engineering 8vo,  4  00 

Wood's  Turbines 8vo,  2  50 


MATERIALS   OF    ENGINEERING. 

Baker's  Roads  and  Pavements 8vo,  5  00 

Treatise  on  Masonry  Construction 8vo,  5  00 

Black's  United  States  Public  Works Oblong  4to,  5  00 

Blanchard's  Bituminous  Roads.      (In  Preparation.) 

Bleininger's  Manufacture  of  Hydraulic  Cement.      (In  Preparation.) 

*  Bovey's  Strength  of  Materials  and  Theory  of  Structures 8vo,  7  50 

Burr's  Elasticity  and  Resistance  of  the  Materials  of  Engineering 8vo,  7  50 

Byrne's  Highway  Construction 8vo,  5  00 

Inspection  of  the  Materials  and  Workmanship  Employed  in  Construction. 

16mo,  3  00 

Church's  Mechanics  of  Engineering 8vo,  6  00 

Du  Bois's  Mechanics  of  Engineering. 

Vol.     I.  Kinematics,  Statics,  Kinetics Small  4to,  7  50 

Vol.  II.  The  Stresses  in  Framed  Structures,  Strength  of  Materials  and 

Theory  of  Flexures Small  4to,  10  00 

*  Eckel's  Cements,  Limes,  and  Plasters 8vo,  6  00 

Stone  and  Clay  Products  used  in  Engineering.      (In  Preparation.) 

Fowler's  Ordinary  Foundations 8vo,  3  50 

*  Greene's  Structural  Mechanics 8vo,  2  50 

*  Holley's  Lead  and  Zinc  Pigments Large  12mo,  3  00 

Holley  and  Ladd's  Analysis  of  Mixed  Paints,  Color  Pigments  and  Varnishes. 

Large  12mo,  2  50 

*  Hubbard's  Dust  Preventives  and   Road  Binders 8vo,  3  00 

Johnson's  (C.  M.)  Rapid  Methods  for  the  Chemical  Analysis  of  Special  Steels, 

Steel-making  Alloys  and  Graphite Large  12mo,  3  00 

Johnson's  (J.  B.)  Materials  of  Construction Large  8vo,  6  00 

Keep's  Cast  Iron 8vo,  2  50 

Lanza's  Applied  Mechanics 8vo,  7  50 

Lowe's  Paints  for  Steel  Structures. ..  .                              12mo,  1  00 


Maire's  Modern  Pigments  and  their  Vehicles 12mo,  $2  00 

Maurer's  Technical  Mechanics 8vo,  4  00 

Merrill's  Stones  for  Building  and  Decoration '.  .8vo,  5  00 

Merriman's  Mechanics  of  Materials 8vo,  5  00 

*  Strength  of  Materials 12mo,  1  00 

Metcalf's  Steel.     A  Manual  for  Steel-users 12mo,  2  00 

Morrison's  Highway  Engineering 8vo,  2  50 

Patton's  Practical  Treatise  on  Foundations 8vo,  5  00 

Rice's  Concrete  Block  Manufacture 8vo,  2  00 

Richardson's  Modern  Asphalt  Pavement 8vo,  3  00 

Richey's  Building  Foreman's  Pocket  Book  and  Ready  Reference.  16mo,mor.  5  00 

*  Cement  Workers'  and  Plasterers'  Edition  (Building  Mechanics'  Ready 

Reference  Series) 16mo,  mor.  1   50 

Handbook  for  Superintendents  of  Construction 16mo,  mor.  4  00 

*  Stone    and     Brick     Masons'     Edition    (Building     Mechanics'    Ready 

Reference  Series) 16mo,  mor.  1   50 

*  Ries's  Clays:   Their  Occurrence,  Properties,  and  Uses 8vo,  5  00 

*  Ries  and  Leighton's  History  of  the  Clay-working  Industry  of  the  United 

States 8vo.  2  50 

Sabin's  Industrial  and  Artistic  Technology  of  Paint  and  Varnish 8vo,  3  00 

*  Smith's  Strength  of  Material 12mo  1   25 

Snow's  Principal  Species  of  Wood 8vo,  3  50 

Spalding's  Hydraulic  Cement 12mo,  2  00 

Text-book  on  Roads  and  Pavements 12mo,  2  00 

*'Taylor  and  Thompson's  Extracts  on  Reinforced  Concrete  Designs.  .  .  .8vo,  2  50 

Treatise  on  Concrete,  Plain  and  Reinforced 8vo,  5  00 

Thurston's  Materials  of  Engineering.      In  Three  Parts 8vo,  8  00 

Part  I.      Non-metallic  Materials  of  Engineering  and  Metallurgy..  .  .8vo,  2  00 

Part  II.     Iron  and  Steel 8vo,  3  50 

Part  III.    A  Treatise  on  Brasses,  Bronzes,  and  Other  Alloys  and  their 

Constituents 8vo,  2  50 

Tillson's  Street  Pavements  and  Paving  Materials 8vo,  4  00 

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Wood's  (De  V.)  Treatise  on  the  Resistance  of  Materials,  and  an  Appendix  on 

the  Preservation  of  Timber 8vo,  2  00 

Wood's  (M.  P.)  Rustless  Coatings:  Corrosion  and  Electrolysis  of  Iron  and 

Steel 8vo,  4  00 


RAILWAY    ENGINEERING. 

Andrews's  Handbook  for  Street  Railway  Engineers 3X5  inches,  mor.  1   25 

Berg's  Buildings  and  Structures  of  American  Railroads 4to,  5  00 

Brooks's  Handbook  of  Street  Railroad  Location IGmc,  mor.  1   50 

Butts's  Civil  Engineer's  Field-book 16mo,  mor.  2  50 

Crandall's  Railway  and  Other  Earthwork  Tables 8vo,  1  50 

Transition  Curve 16mo,  mor.  1  50 

*  Crockett's  Methods  for  Earthwork  Computations 8vo,  1   50 

Dredge's  History  of  the  Pennsylvania  Railroad.    ( 1879) Paper,  5  00 

Fisher's  Table  of  Cubic  Yards Cardboard,  25 

Godwin's  Railroad  Engineers'  Field-book  and  Explorers'  Guide. .  16mo,  mor.  2  50 
Hudson's  Tables  for  Calculating  the  Cubic  Contents  of  Excavations  and  Em- 
bankments    8vo,  1   00 

Ives  and  Hilts's  Problems  in  Surveying,  Railroad  Surveying  and  Geodesy 

16mo,  mor.  1  50 

Molitor  and  Beard's  Manual  for  Resident  Engineers .....  16mo,  1  00 

Nagle's  Field  Manual  for  Railroad  Engineers ....  16mv,  mor.  3  00 

*  Orrock's  Railroad  Structures  and  Estimates 8vo,  3  00 

Philbrick's  Field  Manual  for  Engineers , 16mo,  mor.  3  00 

Raymond's  Railroad  Engineering.     3  volumes. 

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Vol.  HI.  Railroad  Engineer's  Field  Book.      (In  Preparation.) 


Roberts'  Track  Formulae  and  Tables.      (In  Press.) 

Searles's  Field  Engineering IGmo,  mor.  $3  00 

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Railroad  Construction 16mo,  mor.  5  00 

Wellington's  Economic  Theory  of  the  Location  of  Railways Large  12mo,  5  00 

Wilson's  Elements  of  Railroad-Track  and  Construction 12mo,  2  00 


DRAWING. 

Barr's  Kinematics  of  Machinery 8vo,  2  50 

*  Bartlett's  Mechanical  Drawing 8vo,  3  00 

Abridged  Ed Svo,  1   50 

Bartlett  and  Johnson's  Engineering  Descriptive  Geometry.      (In  Press.) 

Coolidge's  Manual  of  Drawing Svo,  paper,  1   00 

Coolidge  and  Freeman's  Elements  of  General  Drafting  for  Mechanical  Engi- 
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Durley's  Kinematics  of  Machines Svo,  4  00 

Emch's  Introduction  to  Projective  Geometry  and  its  Application Svo,  2  50 

Hill's  Text-book  on  Shades  and  Shadows,  and  Perspective Svo,  2  00 

Jamison's  Advanced  Mechanical  Drawing Svo,  2  00 

Elements  of  Mechanical  Drawing . Svo,  2  50 

Jones's  Machine  Design: 

Part  I.    Kinematics  of  Machinery Svo,  1  50 

Part  II.  Form,  Strength,  and  Proportions  of  Parts Svo,  3  00 

*  Kimball  and  Barr's  Machine  Design Svo,  3  00 

MacCord's  Elements  of  Descriptive  Geometry Svo,  3  00 

Kinematics;  or,  Practical  Mechanism Svo,  5  00 

Mechanical  Drawing. 4to,  4  00 

Velocity  Diagrams Svo,  1  50 

McLeod's  Descriptive  Geometry Large  12mo,  1  50 

*  Mahan's  Descriptive  Geometry  and  Stone-cutting Svo,  1  50 

Industrial  Drawing.  (Thompson.) Svo,  3  50 

Moyer's  Descriptive  Geometry Svo,  2  00 

Reed's  Topographical  Drawing  and  Sketching 4to,  5  00 

Reid's  Course  in  Mechanical  Drawing Svo,  2  00 

Text-book  of  Mechanical  Drawing  and  Elementary  Machine  Design.. Svo,  3  00 

Robinson's  Principles  of  Mechanism Svo,  3  00 

Schwamb  and  Merrill's  Elements  of  Mechanism Svo,  3  00 

Smith  (A.  W.)  and  Marx's  Machine  Design Svo,  3  00 

Smith's  (R.  S.)  Manual  of  Topographical  Drawing.  (McMillan.) Svo,  2  50 

*  Titsworth's  Elements  of  Mechanical  Drawing Oblong  Svo,  1   25 

Warren's  Drafting  Instruments  and  Operations 12mo,  1   25 

Elements  of  Descriptive  Geometry,  Shadows,  and  Perspective Svo,  3  50 

Elements  of  Machine  Construction  and  Drawing Svo,  7  50 

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Wilson's  (H.  M.)  Topographic  Surveying Svo,  3  50 

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Woolf's  Elementary  Course  in  Descriptive  Geometry Large  Svo,  3  00 

10 


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Andrews's  Hand-book  for  Street  Railway  Engineering 3X5  inches,  mor.  1  25 

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Anthony  and    Brackett's   Text-book  of  Physics.      (Magie.) ...  .Large  12mo,  3  00 

Benjamin's  History  of  Electricity 8vo,  3  00 

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11 


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12 


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13 


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14 


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15 


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16 


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14  DAY  USE 

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